Binomial Theorem Notes

Binomial Theorem Binomial Expression : Note : If nCx = nCy  Either x = y or x + y = n An expression containing two terms, is called a binomial expression. For example a + b/x, x + 1/ 2 y, a – y etc. are binomial expressions. Expansion n of (x + a) is called Binomial Theorem. Expression containing three terms is called trinomial. For example x + y + z. And in general expression containing more than two terms is called multinomial. Terminology used in binomial theorem : Factorial notation : or n! is pronounced as factorial n and is defined as n! = n(n  1)(n  2)........ 3 . 2.1 ; if n  N  1 ; if n  0  Note : n! = n . (n – 1)! ; from n distinct things mathematically, nCr = n! , (n  r )! r! n N, r  W, 0 r n Properties related to nCr : (i) nCr = nCn – r (iii) n Cr Cr 1 (iv) nCr = = nr 1 r n r n–1 n(n  1) Cr–1 = r(r  1) n–2 Cr–2 n(n  1)(n  2)......... (n  (r  1)) r (r  1)(r  2)....... 2 . 1 = ............. = (v) If n and r are relatively prime, then nCr is divisible by n. But converse is not necessarily true. Statement of binomial theorem : where n  N Mathematical meaning of nCr : The term nCr denotes number of combinations of r things choosen Note : Other symbols of of nCr are n (a + b)n = nC0 anb0 + nC1 an–1 b1 + nC2 an–2 b2 +...+ n Cr an–r br +...... + nCn a0 bn nN n   r  (ii) nCr + nCr – 1 = n + 1Cr and C(n, r). n n or (a + b) =  n Cr anr b r r 0 Note : If we put a = 1 and b = x in the above binomial expansion, then or (1 + x)n = nC0 + nC1 x + nC2 x2 +... + nCr xr +...+ n Cn x n n or (1 + x)n =  n Cr x r r 0 1 Binomial Theorem Regarding Pascal’s Triangle, we note the following : (a) Each row of the triangle begins with 1 and ends with 1. Ex.1 Expand the following binomials : (i) (x – 3)5  3x 2  1    2   (ii) (b) Any entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right. 4 Sol. (i) (x – 3)5 = 5C0x5 + 5C1x4 (– 3)1 + 5C2 x3 (– 3)2 + 5 C3 x2 (–3)3 + 5C4 x (– 3)4 + 5C5 (– 3)5 5 4 3 Ex.2 : The number of dissimilar terms in the expansion of (1 – 3x + 3x2 – x3)20 is 2 = x – 15x + 90x – 270x + 405x – 243 4  3x 2  (ii) 1  2  = 4C0 + 4C1   3  3x 2   2    + 4C  2   3x 2   3x 2  + C3   2  + 4C4   2       3x 2     2    2 4 4 = 1 – 6x2 + 81 8 27 4 27 6 x – x + x 16 2 2 Observations : (i) The number of terms in the binomial expansion (a + b)n is n + 1. (ii) The sum of the indices of a and b in each term is n. (A) 21 (B) 31 2 (C) 41 3 20 (D) 61 3 20 Sol. (1 – 3x + 3x – x ) = [(1 – x) ] = (1 – x)60 Therefore number of dissimilar terms in the expansion of (1 – 3x + 3x2 – x3)20 is 61. General term : (x + y)n = nC0 xn y0 + nC1 xn–1 y1 + ...........+ nCr xn– r r y + ..........+ nCn x0 yn (r + 1)th term is called general term and denoted by Tr+1. Tr+1 = nCr xn–r yr & Tr 1 (n  r  1)  y     Tr r x (iii) The binomial coefficients (nC0, nC1 ..........nCn) Note : of the terms equidistant from the beginning and the The rth term from the end is equal to the (n – r + 2)th end are equal, i.e. nC0 = nCn, nC1 = nCn–1 etc. term from the begining, i.e. nCn – r + 1 xr – 1 yn – r + 1 { nCr = nCn–r} (iv) The binomial coefficient can be remembered with the help of the following pascal’s Triangle (also known as Meru Prastra provided by Pingla) Ex.3 Find (i) 28th term of (5x + 8y)30  4x 5  (ii) 7 term of  5  2x    9 th Index of the binomial The binomial coefficient Sol. (i) T27 + 1 = 30C27 (5x)30– 27 (8y)27 = 30 ! (5x)3 . (8y)27 3 ! 27 !  4x 5  (ii) 7th term of  5  2x    T6 + 1  4x  = C6  5    9! 9  4x  3 9 6 9  5     2x   5  6 = 3 ! 6 !  5   2x  =     6 10500 x3 2 Binomial Theorem n Ex.4 The 7th term from the end in the expansion of [1] 10 10 4 n [2] C42 3 [3] – C32 x n [4] None of these 4 T5 = C4x  2  10 4  1    2  = C .2  2  4 x  x   Ex.5. The ratio of the coefficient of x to the term independent of x in (2) 1 : 32 (3) 32 : 12 (4) 32 : 1 15 15 Ratio is 1 : 32. ( C5 = C10) Ex.6. The term independent of x in the expansion of   is equal to 9 9 9    x 9  3r 0 6 9 r r = Cr (–1) x  r=3 1000  r r and must be integers 2 2 15 [4] none of these 1 9 9 r 6 Sol. [1] Tr + 1 = Cr ( x )   3  r  1 1000 r r  8 6  1000 2 2 = C 3 2   r   The possible set of values of r is {0, 2, 4, ......, 1000} Hence, number of rational terms is 501 [2] C4 [3] C2 Now, ......(vi) The above term will be rational if exponent of 3 and 2 are integers It means 9 [1] – C3 9 ......(v)  1  4 1000 Tr+1 = Cr  9    Coefficient of constant term is = Cr (–1) ......(iv) 1000  r 15 9 r r  r x 6 3 135 30 Sol. The general term in the expansion of 91/ 4  81/ 6 1000 is i.e., Coefficient of x is 6 1  x  3 x  C1a b  Eliminating a and b from (v) and (vi)  n=5 Ex.8 Find the number of rational terms in the expansion of (91/4 + 81/6)1000. Sol. [2] General term in the expansion is Cr x 2 2 n 2 n(n  1) b 10 2 .  9 n(n  1) (n  3) a 3 3 .2 is (1) 12 : 32 C 2ab n 1 ......(iii) Dividing (ii) and (iii) 15 15 10 n b 9  n a 2 (n  1) 2 b 9  (n  1) a 4 Sol.[1] The 7th term from the end = 5th term from beginning 6 ......(i) ......(ii) = 3 Dividing (i) by (ii)  1  C4 24  2  x  10 3 n–3 T4 = C3a b is equal to- 10 Sol. [4] T2 = C1 ab = 135 n 2 n–2 T3 = C2a b = 30 n 10 2   x  2  x   n–1 Ex.9 32 –17 Find the coefficient of x and x  4 1 in  x  3  . x   Sol. Let (r + 1)th term contains xm 9 T = – C3 Ex.7 If the second, third and fourth terms in the n expansion of (a + b) are 135, 30 and 10/3 respectively, then[1] n = 3 [2] n = 2 [3] n = 7 [4] n = 5 r 1  = 15Cr x60 – 7r (– 1)r  x3   Tr + 1 = 15Cr (x4)15 – r    9  3r     6  (i) for x32 , 60 – 7r = 32 7r = 28  r = 4, so 5th term. T5 = 15C4 x32 (– 1)4 Hence, coefficient of x32 is 1365 (ii) for x–17, 60 – 7r = – 17  r = 11 , so 12th term. T12= 15C11 x–17 (– 1)11 Hence, coefficient of x–17 is – 1365 3 Binomial Theorem Middle term(s) : 1 Ex.11 If the middle term in the expansion of x 2 x (a) If n is even, there is only one middle term, which is 6 th is , then 924 x n n 2   term. [1] 10 [2] 12 2  n  (b) If n is odd, there are two middle terms, which are  n  1    2  th  n 1   1  2  [3] 14 [4] None of these th and  th n  Sol. [2] Since n is even therefore  2  1 term is middle   terms. n/2 Note : In any binomial expansion, the middle term(s) has n 2 n/2  1  6 term, hence Cn/2 (x )  x  = 924x greatest binomial coefficient.   n/2 6 In the expansion of (a + b)n  x =x  n = 12. If n No. of greatest Greatest binomial Ex. 12 The greatest coefficient in the expansion of binomial coefficient is coefficient n n Even 1 C n/2 [1] [2] 2n C n! n n Odd 2 C(n – 1)/2 and C(n + 1)/2 [3] 2n C [4] None of these (Values of both these coefficients are equal ) Sol. [1] The greatest coefficient = the coefficient of the middle term Ex.10 Find the middle term(s) in the expansion of 14  x2   2  (i) 1  2    Sol. (i) 1   x 2 a3  (ii)  3a  6    2n Cn n = n! To Determine a Particular Term in the 14     Expansion Here, n is even, therefore middle term is  14  2  th    2  term. It means T8 is middle term  x2  429 T8 = 14C7    = – 16 x14. m , if x occurs in Tr+1 , then r is given by Thus in above expansion if constant term i.e., the term which is independent of x, occurs in Tr+1 then r is determined by 2   In the expansion of n 7  a 3   3 a  (ii)  6   = 9 9  9  1 Here, n is odd therefore, middle terms are  2  th   9 1  &  2  1 th.   It means T5 & T6 are middle terms T5 = C4 (3a)   = 189 a17  8  T6 = 9C5 (3a)9 – 5  a3   6    = – 21 a19.  16  9–4 is – 4  a3   6  9 Ex.13 The term independent of x in the expansion of 5 (1) 5 (3) 7 th th (2) 6 (4) 8 th th 4 Binomial Theorem Sol. Here comparing with Note : If We get  = 2 ,  =1 , n = 9 r= 9(2) =6 n 1 a 1 x itself is a natural number, then Tr = Tr + 1 and both the terms are numerically greatest. th ( 6 + 1) = 7 term is independent of x . 39 Ex.14 The co-efficient of x in the expansion of 4 3 15 ( x – 1/x ) is (1) 455 (2) –455 (3) 105 (4) None of these Sol. From above formula Ex.15 If the sum of the coefficients in expansion of n (1 + 2x) is 6561, the greatest term in the expansion for x [1] 4 15C The required term = T4 = 39 = – 455 x 39 coefficient of x = – 455 4 12 3 3 3 (x ) ( –1/x ) th is [2] 5 th [3] 6 th th [4] 7 Sol. [2] n Sum of the coefficient in the expansion of (1 + 2x) = 6561 , when x = 1 Numerically greatest term in the expansion of n n 8 3 = 6561 3 =3 n=8 (a + b)n, n Î N r 8 Binomial expansion of (a + b)n is as follows : – T Now, (a + b)n = nC0 anb0 + nC1 an–1 b1 + nC2 an–2 b2 +...+ n Cr an–r br +...... + nCn a0 bn  If we put certain values of a and b in RHS, then each term of binomial expansion will have certain T value. The term having numerically greatest value is  said to be numerically greatest term. th Hence, 5 term is the greatest term. Let T r and T r+1 be the rth and (r + 1)th terms respectively Ex.16 The greatest term in the expansion of n n–(r–1) r–1 Tr = Cr–1 a b (1  2 )12 is n n–r r Tr+1 = Cr a b [1] 7th [2] 8th [3] 6th [4] 5th n Tr 1 Now, T r = n Consider Tr 1 Tr  n  r  1   r   b a n 1 –1 r  a b Cr an  r br Cr 1 a nr 1br 1 = n r 1 b . a r Sol. [2] Tr  1 12  r  1  ( 2)  1 Tr r 1 1 13 – r  n 1 r  1 a b    2  1 13   2  r    r  13 2 ( 2 –1) r 2  13 2 r   2  1    r  13 (2 – 2 ) So greatest term will be Tr+1 where r = r  13 (0.586) r  7.616  [ . ] denotes greatest integer function. r=7 So, 8th term is greatest. 5 Binomial Theorem Ex.17 Find the numerically greatest term in the 1 expansion of (3 – 5x)15 when x = 5 . Sol. Let rth and (r + 1)th be two consecutive terms in the expansion of (3 – 5x)15 Tr + 1  Tr 15 Cr 315 – r (| – 5x|)r  15Cr – 1 315 – (r – 1) (|– 5x|)r – 1 15 )! 3. 15 )! |– 5x |  (16  r ) ! (r  1) ! (15  r ) ! r ! 1 5 5. Sol. Let ( 3 + 1)2n = (4 + 2 3 )n = 2n (2 + 3 )n =+f ..........(i) where  and f are its integral & fractional parts respectively 0 < f < 1. Now 0< 3 –1<1 0 < ( 3 – 1)2n < 1 (16 – r) 3r Let 16 – r  3r r4 Result :  n If ( A  B) =  + f, where  and n are positive integers, n being odd and 0 < f < 1, then ( + f) f = kn where A – B2 = k > 0 and A – B < 1. If n is an even integer, then ( + f) (1 – f) = kn Ex.18 If n is positive integer, then prove that the integral part of (7 + 4 3 )n is an odd number.. Sol. Let (7 + 4 3 )n =  + f .............(i) where  & f are its integral and fractional parts respectively. It means 0 < f < 1 Now, 0<7–4 3 <1 0 < (7 – 4 3 )n < 1 Let (7 – 4 3 )n = f  0 < f < 1 .............(ii) Adding (i) and (ii)  + f + f = (7 + 4 3 )n + (7 – 4 3 )n = 2 [nC0 7n + nC2 7n – 2 (4 3 )2 + ..........]  + f + f ( 3 – 1)2n = (4 – 2 3 )n = 2n (2 – 3 )n = f. 0 < f < 1 adding (i) and (ii) 4r  16  Ex.19 Show that the integer just above ( 3 + 1)2n is divisible by 2n + 1 for all n  N. = even integer (f + f must be an integer) 0 < f + f < 2  f + f = 1  + 1 = even integer therefore  is an odd integer. ........(ii) + f + f = ( 3 + 1)2n + ( 3 – 1)2n = 2n [(2 + 3 )n + (2 – 3 )n] = 2.2n [nC0 2n + nC2 2n – 2 ( 3 )2 + ........]  + f + f =2n + 1 k (where k is a positive integer) 0 < f + f < 2  f + f= 1 n+1 +1=2 k.  + 1 is the integer just above ( 3 + 1)2n and which is divisible by 2n + 1. Ex20. Show that 9n + 7 is divisible by 8, where n is a positive integer. Sol. 9n + 7 = (1 + 8)n + 7 = nC0 + nC1 . 8 + nC2 . 82 + ....... + nCn 8n + 7. = 8. C1 + 82. C2 + ....... + Cn . 8n + 8. = 8, where  is a positive integer Hence, 9n + 7 is divisible by 8. Ex.21 What is the remainder when 599 is divided by 13. Sol. 599 = 5.598 = 5. (25)49 = 5 (26 – 1)49 = 5 [49C0 (26)49 – 49C1 (26)48 + .......... + 49C48 (26)1 – 49C49 (26)0] = 5 [49C0 (26)49 – 49C1 (26)48 + ...........+ 49C48 (26)1 – 1] = 5 [49C0 (26)49 – 49C1(26)48 + .......... + 49C48 (26)1 – 13] + 60 = 13 (k) + 52 + 8 (where k is a positive integer) = 13 (k + 4) + 8 Hence, remainder is 8. 6 Binomial Theorem 10 Ex.22 Find the last two digits of the number (17) . 10 5 Sol. (17) = (289) = (290 – 1) Binomial Coefficient & Their Properties 5 n = 5C0 (290)5 – 5C1 (290)4 + ........ + 5C4 (290)1 – 5 C5 (290)0 = 5C0 (290)5 – 5C1 . (290)4 + .........5C3 (290)2 + 5 × 290 – 1 = A multiple of 1000 + 1449 * Hence, last two digits are 49 Note : We can also conclude that last three digits are 449. Ex.23 Which number is larger (1.01)1000000 or 10,000 ? Where C0 = 1, C1 = n, C2 = * Sol. By Binomial Theorem (1.01)1000000 = (1 + 0.01)1000000 * = 1 + 1000000C1 (0.01) + other positive terms * = 1 + 1000000 × 0.01 + other positive terms * = 1 + 10000 + other positive terms Hence (1.01)1000000 > 10,000 Some standard expansions : (i) Consider the expansion n n (x + y) =  r 0 n C r xn–r yr = nC xn y0 + nC xn–1 y1 0 1 + ...........+ nCr xn–r yr + ..........+ nCn x0 yn ....(i) (ii) Now replace y  – y we get n n (x – y) =  Cr (– 1) r xn–r yr = nC0 xn y0 – nC1 xn– n r 0 1 1 n r n–r r n n 0 n y + ...+ Cr (–1) x y + ...+ Cn (– 1) x y ....(ii) (iii) Adding (i) & (ii), we get (x + y)n + (x – y)n = 2[nC0 xn y0 + nC2 xn – 2 y2 +.........] (iv) Subtracting (ii) from (i), we get (x + y)n – (x – y)n = 2[nC1 xn – 1 y1 + nC3 xn – 3 y3 +.........] n * n In the expansion of (1 + x) ; i.e. (1 + x) = C0 + n n r n n C1x +.........+ Cr x +.......+ Cn x n n n The coefficients C0, C1, Cn of various powers of x, are called binomial coefficients and they are written as C0, C1, C2,......,Cn Hence n 2 r (1 + x) = C0 + C1x + C2x +........+ Crx + .......+ n Cnx .....(1) Cr  n(n  1) 2! n(n  1)......... .(n  r  1) , Cn= 1 r! Now, we shall obtain some important expressions involving binomial coefficientssum of coefficient : putting x = 1 in (1), we get n C0 + C1 + C2 +............+ Cn = 2 .....(2) Sum of coefficients with alternate signs : putting x = – 1in (1) We get C0 – C1 + C2 – C3 +...........= 0 .....(3) Sum of coefficients of even and odd terms : from (3) , we have C0 + C2 + C4 +.............= C1 + C3 + C5 +..........(4) i.e. sum of coefficients of even and odd terms are equal. from (2) and (4) n–1  C0 + C2 +.........= C1 + C3 +..........= 2 Sum of products of coefficients : Replacing x by 1/x in (1) We get n 1 C C C  1    C0  1  22  ......... nn  ......... .....(5) x x x x  Multiplying (1) by (5), we get (1  x )2n x n (C 0   (C 0 + C1 C 2  2  .......) x x C 1x + C 2x 2 +........) r Now, comparing coefficients of x on both the sides, we get 2n C0Cr + C1Cr+1 +.........+ Cn–r Cn = Cn–r 2n! = (n  r )! (n  r )! .....(6) 7 Binomial Theorem * Sum of squares of coefficients : putting r = 0 in (6) * we get C0 + C1 + C2 +........+ Cn = n! n! putting r = 1 in (6), we get 2 2 2 2 2n! 2n C0 C1 + C1 C2 + C2 C3 +..........+ Cn–1 Cn = Cn–1 2n! * = (n  1)!(n  1)! .....(7) putting r = 2 in (6), we get 2n C0 C2 + C1 C3 + C2 C4 +..........+ Cn-2 Cn = Cn–2 2n! = (n  2 )! (n  2 )! * 2 n–1 n(1+ x) = C1 + 2C2x + 3C3x +.........+ nCn x Now putting x = 1 and x = –1 respectively n–1 C1 + 2C2 + 3C3 +........+ nCn = n.2 * * .....(9) .....(10) adding (2) and (9) (n+1) n–1 C0 + 2C1 + 3C2 +..........+ Cn = 2 (n+2) .....(11) Integrating (1) w.r.t. x between the limits 0 to 1, we get, 1 1  (1  x)n 1   x2 x3 C x n 1   C2  ........  n    C0 x  C1  2 3 n  1   n  1  0  0  C0+ * C1 C2 C 2n 1  1   .........  n  2 3 n 1 n 1 ....(12) Integrating (1) w.r.t. x between the limits -1 to 0, we get, 0 0  (1  x)n 1   x2 x3 C x n1   C2  ........  n    C0 x  C1  2 3 n  1   n  1  1  1 n  C0– =  (r  1) . C1 C2 C3 (1) .Cn 1    .........  2 3 4 n 1 (n 1) ....(13) n n–1 (ii) C0 + 2C1 + 3. C2 + ........ + (n + 1) Cn = 2 (n + 2). (iii) C0 – C C1 C C 1 + 2 – 3 + ..... + ( –1)n n = . 3 2 4 n 1 n 1 r 0  Cr 1 +  n Cr r 0 n Cr = n . 2n – 1 + 2n = 2n – 1 (n + 2). r 0 C C1 C + 32 – 3 + ........ + (– 1)n. 2 4 n Cn n 1 = = n (i) C0 + 3C1 + 3 C2 + .......... + 3 Cn = 4 .  n n 1 L.H.S. = C0 – n Ex.24 If (1 + x) = C0 + C1x + C2x + ............. + cnx , then show that n r. nCr + RHS II. Method : By Differentiation (1 + x)n = C0 + C1x + C2x2 + ........... + Cnxn Multiplying both sides by x, x(1 + x)n = C0x + C1x2 + C2x3 + ........ + Cn xn + 1. Differentiating both sides (1 + x)n + x n (1 + x)n – 1 = C0 + 2. C1x + 3 . C2x2 + ....... + (n + 1)Cnxn. putting x = 1, we get C0 + 2.C1 + 3 . C2 + ...... + (n + 1) Cn = 2n + n . 2n – 1 C0 + 2.C1 + 3 . C2 + ...... + (n + 1) Cn = 2n – 1 (n + 2) Proved (iii) I Method : By Summation 1 n 1 =  (1) n r . r 0 n  (1) r . r 0 Cr r 1 n 1  n Cr + 1  r  1 . Cr   n+ 1  C r 1   n 1 1 [n + 1C1 – n + 1C2 + n + 1C3 – .............+ (– 1)n . n1 n+1 2 2 Cr = r 0 n = n n r 0 Differentiating both sides of (1) w.r.t. x, we get and C1 – 2C2 + 3C3–.........= 0 n n =n  .....(8) n–1 Sol. (i) (1 + x)n = C0 + C1 x + C2x2 + ........... + Cnxn put x = 3 C0 + 3 . C1 + 32 . C2 + .......... + 3n . Cn = 4n (ii) I. Method : By Summation L.H.S. = nC0 + 2. nC1 + 3 . nC2 + ........ + (n + 1). nCn. Cn + 1] 1 [– n1 n+1 C0 + n + 1C1 – n + 1C2 + ......... + (– 1)n . n + 1Cn + 1 + n + 1C0] = 1 = R.H.S. , since n 1  n1 C0  n1 C1 n 1 C2  ...  ( 1)n n1 Cn1  0 8 Binomial Theorem Ex.27 If (1 + x) = C0 + C1x + C2x2 + ........+ Cnxn, then prove that n II Method : By Integration (1 + x)n = C0 + C1x + C2x2 + ...... + Cn xn. Integrating both sides, within the limits – 1 to 0. (i) C02 + C12 + C22 + ...... + Cn2 = 2nCn 0 (ii) C0 C2 + C 1C3 + C2C4 + .......... + C n – 2 C n = 2nCn – 2 or 2nCn + 2  (1  x )n  1     n  1  1 (iii) 1. C02 + 3 . C12 + 5. C22 + ......... + (2n + 1) . Cn2 . = 2n. 2n – 1Cn + 2nCn. 0  x2 x3 x n1  = C 0 x  C1 2  C 2 3  .....  C n n  1   1 Sol. 1 –0=0– n 1 C1 C 2  n 1 C n    C 0  2  3  .....  ( 1) n  1   C C C 1 C0 – 1 + 2 – .......... + (– 1)n n = 3 2 n1 n1 (i) (1 + x)n = C0 + C1x + C2x2 + ......... + Cn xn. .......(i) (x + 1)n = C0xn + C1xn – 1+ C2xn – 2 + ..... + Cn x0....(ii) Multiplying (i) and (ii) Proved Ex.25 If C 0 , C 1 , C 2 , ..............C n are binomial coefficients then 1 1 1 1    .....  n ! 0 ! (n  1) ! (n  2) ! 2 ! 0 ! n ! is equal to2 n1 [2] n ! n [1] 2 2n n (divide it by n !) Ex.26 If C0, C1, C2 .........Cn are binomial coefficients, 2n 1  1 n 1 2 n1 [3] n1 is equal to- [2] (ii) From the product of (i) and (ii) comparing coefficients of xn – 2 or xn + 2 both sides, 2 L.H.S. = 1. C02 + 3. C12 + 5. C22 + .......... + (2n + 1) Cn2. n! n! n! n! 2    .....  = (n  1)! (n  1) ! 1 ! (n  1) ! 2 ! 0!n! n! [1] C02 + C12 + C22 + ........ + Cn2 = 2nCn (iii) I Method : By Summation n then C0 + Comparing coefficient of xn, C0C2 + C1C3 + C2C4 + ........ + Cn – 2 Cn = 2n C n – or 2nCn + 2. [3] n ! [4] none of these Sol. [3] n C0 + C1 + C2 + ....... + Cn = 2 C1 C 2 C   ...... n 2 3 n 1 (C0 + C1x + C2x2 + ......... + Cnxn) (C0xn + C1xn – 1 + ......... + Cnx0) = (1 + x)2n 2 n1  1 n 1 [4] none of these Sol. [2] n 2 n (1 + x) = C0 + C1x + C2x + .....+Cnx Integrate it with respect to x C x2 (1  x )n1 x3 x n 1  C0 x  1  C 2  .....  C n K n1 2 3 n1 1 Put x = 0 then, K = , Now put x = 1 n 1 = n n  (2r  1) C n 2 r r0 =  2.r . (nCr)2 + r 0 ( n Cr ) 2 r0 n =2 . n . n– 1 Cr – 1 nCr + 2nCn r 1 n (1 + x) = nC0 + nC1 x + nC2 x2 + .............nCn xn ..........(i) (x + 1)n – 1 = n – 1C0 xn – 1 + n–1 Cn – 1x0 .........(ii) n–1 C1 xn – 2 + .........+ Multiplying (i) and (ii) and comparing coeffcients of xn. n–1 C0 . nC1 + n – 1C1 . nC2 + ........... + n – 1Cn – 1 . nCn = 2n – 1Cn n  n 1 Cr 1 . nCr = 2n – 1 Cn r 0 Hence, required summation is 2n. 2n – 1Cn + 2nCn = R.H.S. 9 Binomial Theorem II Method : By Differentiation Sol. [2] n (1 + x2)n = C0 + C1x2 + C2x4 + C3x6 + ..............+ Cn x2n x(1 + x2)n = C0x + C1x3 + C2x5 + ............. + Cnx2n + 1. Differentiating both sides x . n (1 + x2)n – 1 . 2x + (1 + x2)n = C0 + 3. C1x2 + 5. C2 x4 + ....... + (2n + 1) Cn x2n ........(i) (x2 + 1)n = C0 x2n + C1 x2n – 2 + C2 x2n – 4 + ......... + Cn ........(ii) Multiplying (i) & (ii) (C0 + 3C1x2 + 5C2x4 + ......... + (2n + 1) C n x2n) (C0 x2n + C1x2n – 2 + ........... + Cn) = 2n x2 (1 + x2)2n – 1 + (1 + x2)2n 2 2 n C02 + 3C12 + 5C22 + .........+ (2n + 1) Cn2 = 2n . 2n–1 Cn + 2nCn. Proved Ex.28 The value of n n n 3 C0 – 8 C1 + 13 C2 – 18 C3 + ......... + n terms is n [1] 0 [2] 3 n [3] 5 n n (1  x )m 1  x 3(0) – 5 (0) = 0 n Ex.29 If C r stands for C r , then the value of 2C0 – n 1 23 n+1 2 22 C1 + C2 + ...... + (–1) (n  1) 3 2 is equal to[1] 0, if n is even 1 , if n is even n1 nm1 [2] 0, if n is odd [4] 1 , if n is odd n1 1 = x n 3 C 0 – 8 C 1 + 13 C 2 – 18 C 3 + ...... n n n n n n 3 ( C0 – C1 + C2 – C3 + .......) – 5 ( C1 –2 C2 + n 3 C3 –.........) [3] = m+3Cm+1 + ............. + nCm = nCm+1 + nCm = n+1Cm+1 II Method m Cm + m+1Cm + m+2Cm + .......... + nCm The above series can be obtained by writing the coefficient of xm in (1 + x)m + (1 + x)m+1 + ......... + (1 + x)n Let S = (1 + x)m + (1 + x)m+1 +.............. + (1 + x)n = n m 2 C 1  Cm   + .................. + nC = m m [4] none of these Sol. [1]  m 1 C m1 m 1 Cm + m+2C + .............. + nC =  m m    m2 2 C0 + 3C1 + 5C2 + .........+ (2n + 1) C = 2n . 2n – 1 Cn – 1 + 2nCn. n 22 23 (2)n 1  C2  .....  C n 2 3 (n  1) If n is odd, then L.H.S. = 0. Ex.30 Find the summation of the following series – (i) mCm + m+1Cm + m+2Cm + .............. + nCm (ii) nC3 + 2 . n+1C3 + 3. n+2C3 + ......... + n . 2n–1C3 Sol. (i) I Method : Using property, nCr +nCr–1 = n+1Cr m Cm + m+1Cm + m+2Cm + .............. + nCm  2C0  C1 { mCm = m+1Cm+1} comparing coefficient of x2n, 2 n (  1) n  1  1 n 1 Put x = –2 Multiplying both sides by x 2 (1 + x) = C0 + C1x + C2x + ......+ Cnx = coefficient of xm in 1 x x 1 x n 1 – n 1  1 x x m 1 xm = n + 1Cm +1 x + 0 = n + 1Cm +1 (ii) nC3 + 2 . n+1C3 + 3 . n+2C3 + .......... + n . 2n–1C3 The above series can be obatined by writing the coefficient of x3 in (1 + x)n + 2 . (1 + x)n+1 + 3 . (1 + x)n+2 + ........... + n . (1 + x)2n–1 Let S = (1 + x)n + 2 . (1 + x)n+1 + 3. (1 + x)n+2 + ........... + n (1 + x)2n–1 .....(i) n+1 n+2 (1 + x)S = (1 + x) + 2 (1 + x) + ............. + (n – 1) (1 + x)2n–1 + n(1 + x)2n ....(ii) 10 Binomial Theorem Subtracting (ii) from (i) Multinomial theorem : – xS = (1 + x)n + (1 + x)n+1 + (1 + x)n+2 + .............. + (1 + x)2n–1 – n(1 + x)2n = (1  x )n (1  x )n  1 x + 2n  (1  x) n  (1  x ) + x2 n(1  x )2n x 4n 4n 4n 4n–2 [2] 2 2n–1 [3] 2 4n–2 2n–1 +2 [4] 2 n 4n 4n 4n 2 = 4n 3 4n 4 (1 + x) = C0 + C1 x + C2x + C3x + C4x 4n 4n + ........+ Cnx Put x = 1 and x = –1, then adding. 4n–1 2 4n 4n xn–r yr = 4n n! r ! r2 ! n 1  x r1 . yr2 Total number of terms in the expansion of (x + y)n is equal to number of non-negative integral solution of r1 + r2 = n i.e. n+2–1C2–1 = n+1C1 = n + 1 In the same fashion we can write the multinomial theorem (x1 + x2 + x3 + ........... xk)n 4n–2 (–1) + 2 Sol. [4] 4n n!  (n  r )! r! r1  r2 Ex.31 The value of C0 + C4 + C8 + ........+ C4n is[1] 2 = r 0 therefore, (x + y)n n(1  x )2n x 4n Cr xn–r yr r 0 Hence, required summation of the series is – 2nC5 + n C5 + n . 2nC4 4n n putting n – r = r1 , r = r2 (coefficient of x in S) 2n  (x + y)n = 3 x :S x3 : – n (1 + x) x2 3 n n  (1  x )2n  (1  x )n S= As we know the Binomial Theorem –  r1  r2 ... rk n! x r1 . x r22 ...x rkk r ! r !... rk ! 1 n 1 2 Here total number of terms in the expansion of (x1 + x2 + .......... + xk)n is equal to number of nonnegative integral solution of r1 + r2 + ........ + rk = n i.e. n+k–1Ck–1 4n = C0 + C2 + C4 + ....... + C4n ......(i) Now put, x = i (1 + i) = C0 + C1i – C2 – C3i + C4 + .......+ 4n C4n Ex.33 Find the coefficient of a2 b3 c4 d in the expansion of (a – b – c + d)10 Compare real and imaginary part, we get Sol. (a – b – c + d) 10 4n n 4n 2n 4n 4n 4n 4n 4n 4n 4n 4n 4n (–1) (2) = C0 – C2 + C2 + C4 – C6 + 4n ....+ C4n ......(ii) Adding (i) and (ii), we get  4n 4n 4n n 2n–1 C0 + C4 + ..... + C4n = (–1) (2) 4n–2 +2 Ex.32 aC0 + (a + b) C1 + (a + 2b) C2 + .......+ (a + nb) Cn is equal ton [1] (2a + nb) 2 [2] (2a + nb) 2 n [3] (na + 2b) 2 [4] (na + 2b) 2  = r1 r2  r3  r4 r1 r2 r3 (a) ( b) ( c ) (d) r4 we want to get a2 b3 c4 d this implies that r1 = 2, r2 = 3, r3 = 4, r4 = 1 (10 )! coeff. of a2 b3 c4 d is 2! 3! 4! 1! (–1)3 (–1)4 = – 12600  n–1 n–1 (10 )! r ! r !r ! r ! 10 1 2 3 4 11 7 Ex.34 In the expansion of 1  x   , find the term  x independent of x. Sol. [2] a (C0 + C1 + C2 + ......... + Cn) + b(C1 + 2C2 + n ...... + Cn) n n–1 = a.2 + b. n 2 =2 n  2a  nb  n–1   = (2a + nb) 2 2   11  7   Sol. 1  x  x  =  r1  r2  r3 (11)! r !r !r ! 11 1 2 3 r 73 (1)r1 ( x )r2   x The exponent 11 is to be divided among the base variables 1, x and 7 in such a way so that we x get x0. 11 Therefore, possible set of values of (r1, r2, r3) are (11, 0, 0), (9, 1, 1), (7, 2, 2), (5, 3, 3), (3, 4, 4), 2. (1, 5, 5) (a) (1 + x) = 1 + n x + If x < 1 and n Q but n N, then n n(n  1) 2 n(n  1)......(n  r  1) r x  ......  x  .... 2! r! Hence the required term is (11)! (11)! (11)! (11)! 0 1 2 3 (7 ) + 7 + 7 + 9! 1 !1! 7! 2 ! 2 ! 5! 3 ! 3 ! 7 (11)! (11)! (11)! + 3! 4 ! 4 ! 74 + 1 ! 5 ! 5 ! 75 (11)! 2! (11)! 4! (11) ! = 1 + 9 ! 2 ! . 1 ! 1 ! 71 + 7 ! 4 ! . 2 ! 2 ! 72 + 5 ! 6 ! . 6! (11) ! 8! (11) ! (10) ! 3 4 5 7 + . 7 + . 3!3! 3!8! 4! 4! 1 ! 10 ! 5 ! 5 ! 7 11 2 1 11 4 2 11 5 =1+  11 n (b) (1 – x) = 1 – nx + n ( n  1) 2 n ( n  1)( n  2 ) 3 x  x  .... 2! 3! n(n  1).....(n  r  1) ( x)r  ..... r! –n (c) (1 – x) = 1 + nx +  n (n  1) 2 n (n  1)(n  2) 3 x  x  .... 2! 3! 6 = 1 + C2 . C1 . 7 + C4 . C2 . 7 + C6 . C3 . 73 + 11C8 . 8C4 . 74 + 11C10 . 10C5 . 75 C2r . 2rCr . 7r n(n  1).....(n  r 1) r x  ..... r! –n (d) (1 + x) = 1 – nx +  n(n  1) 2 n(n  1)(n  2) 3 x  x  .... 2! 3! r 1 Binomial Theorem for any Index (i) Its first term is 1, and (ii) Its second term is numerically less than 1. Thus when n  N and x < 1, then it states * * n (1 + x) = 1 + nx + n (n  1) 2 n (n  1) (n  2) 3 x  x  ....... 2! 3! n(n  1) (n  r  1) r  x  ......... r! * General Term : Tr 1 n(n  1).....(n  r 1) (x)r  ..... r! By putting n = 1, 2, 3 in the above results (c) and (d), we get the following results–1 2 3 r (1 – x) = 1 + x + x + x + ...... + x + ....... r General term Tr + 1 = x –1 2 3 r (1 + x) = 1 – x + x – x + ...... (–x) + ...... r General term Tr + 1 = (–x) –2 2 3 r (1 – x) = 1 + 2x + 3x + 4x + ..... + (r + 1) x + ..... r General term Tr + 1 = (r + 1) x –2 2 3 (1 + x) = 1 – 2x + 3x – 4x + ....... + (r + 1) r ( –x) + ...... r General term Tr + 1 = (r + 1) (–x) –3 2 3 (1 – x) = 1 + 3x + 6x + 10x + ...... +  When n is a negative integer or a fraction then the expansion of a binomial is possible only when 1. Binomial Theorem Some Important Expansions : * * * n(n  1)(n  2)......... (n  r  1) r  x r! (r  1)(r  2) r x  .... 2! Note : General term  (r  1)(r  2) x r (i) In this expansion the coefficient of different terms * n n n can not be expressed as C0, C1, C2......because n is not a positive integer. (1 + x) (ii) In this case there are infinite terms in the expansion. –3 2! 2 3 = 1 – 3x + 6x – 10x + ......... + (r  1)(r  2) ( x )r  ........ 2! General term  (r  1)(r  2) (  x)r 2! 12 Binomial Theorem Ex.37 Prove that the coefficient of xr in (1 – x)–n is n+r–1Cr Ex.35 If x is very large and n is a negative integer or a proper fraction, then an approximate value of Sol. (r + 1)th term in the expansion of (1 – x)–n can be written as n  1 x    is equal to x  n(n  1)(n  2)......( n  r  1) (–x)r r! Tr +1 = [1] 1 + x n [2] 1 + n x [3] 1 + 1 x [4] n 1  x  = (–1)r  1   Sol. [2] but as x is very large terms therefore after 2nd term other terms can be ignored. Ex.36 If is approximately equal to n(n  1)(n  2)......(n  r  1) (–x)r r! = n(n  1)(n  2)......(n  r  1) r x r! = (n  1)! n(n  1)......(n  r  1) r x (n  1) ! r ! Hence, coefficient of xr is (n  r  1)! n+r–1 = Cr Proved (n  1)! r! Ex.38 If x is so small such that its square and higher powers may be neglected, then find the value of (1  3x)1/ 2  (1  x )5 / 3 for small values of x, then a,b ( 4  x )1/ 2 [1] [2] [3] Sol. [2] [4] Sol. (1  3x)1/ 2  (1  x )5 / 3 ( 4  x )1/ 2 = (1  3 x )1/ 2  (1  x )5 / 3  2 1   x 4  3 5x x  1 2 3 1/ 2 x   21    4 1 1/ 2 x 1  2  19 x   1   = 6   4 2   1   5 1  1 1 52 1 2  2 1 2 (3x)  2  2 2 (3x) ......  1 3(x)  3 3 2(x) ......       2  1 x  1 1 1  x   = 21        ........   2 4 2 2 2  4  = 1 / 2 x 19  1  2  19 x  1  x  1  = 2  4  6 x 6 8     2  2  x 19 41 =1– 8 – x =1– x 12 24 53 2  19  1  12 x  144 x  .........    1  35 x  ......   24  x 1 2  1  2  8 x  ........   Neglecting higher powers of x, then a + bx = 1 – 35 x 24  a = 1, b = – 35 . 24 13