Binomial Theorem

BINOMIAL THEOREM BINOMIAL EXPRESSION Any algebraic expression which contains two dissimilar terms is called Binomial expression. For example : x – y, xy + 1 1 1 ,  1,  3 etc. x z ( x  y )1/3 Terminology Used in Binomial Theorem Factorial notation : or n! is pronounced as factorial n and is defined as  n(n  1)(n  2)........3.2.1 ; if n  N n! =  1 ; if n  0   n! = n . (n – 1)! ; nN BINOMIAL THEOREM The formula by which any positive integral power of a Binomial expression can be expanded in the form of a series is known as BINOMIAL THEOREM. If x, y  R and n  N, then : ( x + y)n = nC0xn + nC1xn-1 y + nC2xn-2 y2 + ..... + nCrxn-r yr + ..... + nCnyn = n  r 0 n Cr x n  r y r This theorem can be proved by induction. (a) The number of terms in the expansion is ( n+1) i.e. one more than the index. (b) The sum of the indices of x & y in each term is n. (c) The Binomial coefficients of the terms (nC0, nC1.....) equidistant from the beginning and the end are equal. i.e. nCr = nCn – r n (d) Symbol nCr can also be denoted by   , C(n, r) or A nr . r  r n n  The coefficient of x in (1+x) = Cr & that in (1–x)n = (–1)r .nCr Some Important Expansions : (i) (1 + x)n = nC0 + nC1x + nC2x2 + ........ + nCnxn. (ii) (1 – x)n = nC0 – nC1x + nC2x2 + ........ + (–1)n . nCnxn. Ex. Expand the following Binomials : (i) Sol. (x – 3) (ii) 5  3x 2  1   2   4 (i) (x – 3)5 = 5C0x5 + 5C1x4 (– 3)1 + 5C2 x3 (– 3)2 + 5C3 x2 (–3)3 + 5C4 x (– 3)4 + 5C5 (– 3)5 = x5 – 15x4 + 90x3 – 270x2 + 405x – 243 4  3x 2   = 4 C 0 + 4C 1 (ii) 1  2   = 1 – 6x2 + 2 3  3x 2   3x 2   3x 2   3x 2    + 4C2    + 4C3    + 4C4     2   2   2   2  27 4 27 6 81 8 x – x + x 2 2 16 4 18 3  73  3.18.7.25  Ex. Find the value of Sol. The numerator is of the form a3 + b3 + 3ab(a + b) = (a + b)3 Where, a = 18 and b = 7  Nr = (18 + 7)3 = (25)3 Denominator can be written as 36  6.243.2  15.81.4  20.27.8  15.9.16  6.3.32  64 3 6  6 C 1 .3 5 .2 1  6 C 2 .3 4 .2 2  6 C 3 .3 3 .2 3  6 C 4 3 2 .2 4  6 C 5 3 .2 5  6 C 6 2 6   3  2   5 6  2 5  6  Nr (25)3  1 Dr (25)3 Pascal's Triangle A triangular arrangement of numbers as shown. The numbers give the coefficients for the expansion of (x + y)n. The first row is for n = 0, the second for n = 1, etc. Each row has 1 as its first and last number. Other numbers are generated by adding the two numbers immediately to the left and right in the row above. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 etc. IMPORTANT TERMS IN THE BINOMIAL EXPANSION (a) General Term : The general term or the ( r +1)th term in the expansion of (x + y)n is given by Tr +1= nCr x n–r yr Ex. Find (i) 28th term of (5x + 8y)30 Sol. (i) T 27 + 1 = 30C27 (5x)30– 27 (8y) 27 = (ii)  4x 5   7th term of    5 2x   4x  T 6 + 1 = 9C 6    5  (b) 96 (ii)  4x 5   7 th term of    5 2x  9 30 ! (5x)3 . (8y)27 3! 27 ! 9  5     2x  6 3 = 9!  4x   5  6 10500 =   3!6 !  5   2x  x3 Middle Term : The middle term(s) in the expansion of (x + y)n is (are) : (i) If n is even, there is only one middle term which is given by T(n +2)/2= nCn/2. x n/2. yn/2 (ii) If n is odd, there are two middle terms which are T(n +1)/2 & T[(n+1)/2]+1 3 BINOMIAL THEOREM Middle term has greatest Binomial coefficient and if there are 2 middle terms their coefficients will be equal.  When r= n if n is even 2 nCr will be maximum n–1 n+1 When r= or if n is odd 2 2  Ex. The term containing greatest Binomial coefficient will be middle term in the expansion of (1 + x)n Find the middle term(s) in the expansion of 14 (i)  x2  1   2   (i)  x2  1   2   (ii)  a3  3a    6  9 14 Sol.  14  2  th Here, n is even, therefore middle term is   term.  2  It means T 8 is middle term 7 429 14  x2  T 8 = C7    = – x . 16  2  14 (ii)  a3  3a    6  9  9  1  th Here, n is odd therefore, middle terms are   &  2  It means T 5 & T 6 are middle terms  9  1  th  1 .   2  4 189 17  a3  T 5 = 9C4 (3a) 9 – 4    = a 8  6 5  a3  21 19  = – 16 a . 6   T 6 = 9C5 (3a) 9 – 5   (c) Term Independent of x : Term independent of x does not contain x ; Hence find the value of r for which the exponent of x is zero. 10 Ex.  x  3    2  Find the term independent of x in   2 x    3 Sol. General term in the expansion is r 10  x 2  3  Cr    2   3   2x  10  r 2 3r  10C r x 2 10 . 35 r 2 10  r 2 . For constant term, 20 3r = 10  r = 3 2 which is not an integer. Therefore, there will be no constant term. (d) Numerically Greatest Term : Binomial expansion of (a + b)n is as follows : – (a + b)n = nC0 anb 0 + nC1 an–1 b 1 + nC2 an–2 b 2 +...+ nCr an–r b r +...... + nCn a0 b n If we put certain values of a and b in RHS, then each term of Binomial expansion will have certain value. The term having numerically greatest value is said to be numerically greatest term. Let T r and T r+1 be the rth and (r + 1)th terms respectively Tr = nCr–1 a n–(r–1) br–1 Tr+1 = nCr a n–r br Cr a n r b r Cr 1 a n r 1br 1 n Tr 1 = Tr Now, n Tr 1 Tr Consider b 1 a a b n 1 –1 r Case -  When (i) n 1 a 1 b T r+1 > T r T r+1 = T r i.e. (iii) n 1 a 1 b is an integer (say m), then i.e. (ii) b n  r 1 . a r 1  n  r 1   r   r = T r+1 < T r i.e. when r T 1, T 3 > T2, ......., T m > T m–1 when r=m T m+1 = T m when r>m (r = m + 1, m + 2, ..........n ) T m+2 < T m+1 , T m+3 < T m+2 , ..........T n+1 < T n Conclusion When n 1 is an integer, say m, then Tm and Tm+1 will be numerically greatest terms (both terms are equal in magnitude) a 1 b Case -  When (i) n 1 is not an integer (Let its integral part be m), then a 1 b T r+1 > T r i.e. when r< n 1 a 1 b (r = 1, 2, 3,........, m–1, m) T 2 > T1 , T 3 > T2, .............., T m+1 > Tm BINOMIAL THEOREM (ii) T r+1 < T r i.e. when r > n 1 a 1 b (r = m + 1, m + 2, ..............n) T m+2 < T m+1 , T m+3 < T m+2 , .............., T n +1 < T n Conclusion When n 1 is not an integer and its integral part is m, then T m+1 will be the numerically greatest term. a 1 b (i) In any Binomial expansion, the middle term(s) has greatest Binomial coefficient. In the expansion of (a + b)n n No. of Greatest Binomial Coefficient Even 1 Odd 2 Greatest Binomial Coefficient Cn/2 n n C(n – 1)/2 and nC(n + 1)/2 (Values of both these coefficients are equal ) (ii) In order to obtain the term having numerically greatest coefficient, put a = b = 1, and proceed as discussed above. Ex. Find numerically greatest term in the expansion of (3 – 5x)11 when x = Sol. Using 1 5 n 1 n 1 1  r  a a 1 1 b b 11  1 11  1 1  r  3 3 1 1 5 x 5 x Solving we get 2 < r < 3 r = 2, 3  So, the greatest terms are T2+1 and T3+1. Greatest term (when r = 2)  11 T3 = C2.39 (–5x)2 = 55.39 = T4 From above we say that the value of both greatest terms are equal. Ex. 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  .............(ii)  0 < f < 1 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  = even integer  (f + f  must be an integer) 01 then we may find it convenient to expand in powers of 1/x, which then will be small. (f) (1 – x)–3 = 1 + 3x + 6x2 + 10x3 + ..... + Approximations n( n  1) 2 n(n  1)(n  2) 3 x+ x ....... 1.2.3 1.2 If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its square and higher powers may be neglected then (1 + x)n = 1 + nx, approximately. This is an approximate value of (1 + x)n (1 + x)n = 1 + nx + Ex. Prove that the coefficient of xr in (1 – x)–n is n+r–1Cr Sol. (r + 1)th term in the expansion of (1 – x)–n can be written as T r +1 = n(n  1)( n  2)......( n  r  1) (–x) r r! = (–1) r = n(n  1)(n  2)......(n  r  1) r n(n  1)(n  2)......(n  r  1) (–x)r = x r! 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! BINOMIAL THEOREM Ex. If x is so small such that its square and higher powers may be neglected then find the approximate value of (1  3 x )1/2  (1  x )5/3  4  x 1/2 Sol. 3 5x 1/2 1 x 1 (1  3 x )1/2  (1  x )5/3 1 x 19  x 19 41 1 19  x  2 3 x 2  x 1  = =   =  2   x  = 1  x = 1 –  1/2  4  x 1/2 2 4 6   8 12 24 2 6  4     x 2 1    4 EXPONENTIAL SERIES (a) e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is 2.7182818284. (b) Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier, their inventor. They are also called Natural Logarithm. (c) ex = 1  x x 2 x3  1    ....... ; where x may be any real or complex number & e = Lim  1   n   n 1! 2! 3! (d) ax = 1  x x2 x3 l na  l n 2 a  l n 3a  ....... , where a > 0 1! 2! 3! (e) e = 1 Ex. Find the coefficient of x2 in the expansion of e2x+3 as a series in powers of x. Sol. In the exponential series n 1 1 1    ....... 1! 2! 3! ex = 1 + x x2 x3   + .... 1! 2! 3!  2x (2x)2 (2x)3     ... Hence e2x+3 = e3 . e2x = e3 1  1! 2! 3!   Thus, the coefficient of x2 in the expansion of e2x+3 is e3 . 22 = 2e3. 2! LOGARITHMIC SERIES 2 3 4 (a) n (1 + x) = x  x  x  x  ....... , where –1 < x  1 2 3 4 where –1  x  1  1  1  1  1  .......  l n 2 2 Ex. 3 4  elnx = x ; for all x > 0  n2 = 0.693 x 2 x3 x 4    ....... , 2 3 4  n10 = 2.303 If ,  are the roots of the equation x2 – px + q = 0, prove that loge (1 + px + qx2) = ( + )x – Sol. (b) n (1 - x) = - x   2   2 2  3  3 3 x + x – ..... 2 3     2 x 2 3 x 3   2 x 2 3 x 3  – ..... +  x –  – .... Right hand side = x – 2 3 2 3     = loge(1 + x) + loge(1 + x) = loge (1 + ( + )x + x2) = loge (1 + px + qx2) = Left hand side. Here, we have used the facts  +  = p and  = q. We know this from the given roots of the quadratic equation. We have also assumed that that both | .x | < 1 and | x | < 1. n If x, y  R and and n  N, then : (x + y)n = nC0xn + nC1xn-1y + nC2xn-2 y2 + ......+ nCrxn-r yr + ....+ nCn yn 1. Important Terms in the Binomial Expansion (a) General term : The general terms or the ( r + 1)th term in the expansion of (x + y)n is given by = Tr + 1 = nCr xn-r.yr (b) Middle Term The middle term (s) is the expansion of ( x + y)n is (are) : (i) If n is even , there is only one middle terms which is given by T(n+2)/2 = nCn/2 . xn/2 . yn/2. (ii) If n is odd, there are two middle terms which are T(n +1)/2 & T[(n +1)/2]+1 (c) Term Independent of x Term independent of x contains no x ; Hence find the value of r for which the exponent of x is zero. 2. If  A B  n = I + f, where I & n are positive integers & 0  f < 1, then (a) (I + f). f = Kn if n is odd & A – B2 = K > 0 (b) (I + f) (1 – f) = kn if n is even & 3. A –B<0 Some Results on Binomial Coefficients (a) nCx = nCy  x = y or x + y = n (b) nCr-1 + nCr = n+1Cr Cn 2 n 1  1  n 1 n 1 (c) C0 + C1 C  2 2 3 (d) C0 –  1 n Cn 1 C1 C2 C3    ......+ n 1 n 1 2 3 4 + ...... (e) C0 + C1 + C2 +........ = Cn = 2n (f) C0 + C2 + C4 +.........= C1 + C3 + C5 + ....... = 2n-1 (g) C02 + C12 + C22 + ....... + Cn2 = 2nCn =  2n ! n!n! (h) C0.Cr + C1.Cr+1 + C2.Cr+2 + .......+ Cn-rCn = Remenber : (2n)! = 2n. n! [1.3.5.........(2n – 1)]  2n !  n  r ! n  r !  r 0 n Cr x n r yr BINOMIAL THEOREM 4. Greatest Coefficient & Greatest Term in Expansion of ( x + a)n (a) If n is even greatest coefficient is nCn/2 n If n is odd greatest coefficient is Cn 1 or 2 n Cn 1 2 (b) For greatest term : n 1  is an integer  Tp & TP 1 if x  1 a  Greatest term =  T if n  1 is non integer and   q, q  1 , q  I  q 1 x 1  a  5. Binomial Theorem for Negative or Fractional Indices If n  Q, then (1 + x)n = 1 + nx + n  n  1 2! x2 + n  n  1 n  2  3! x3 + ......  provided |x| < 1 Note (i) (1 – x)–1 = 1 + x + x2 + x3 + .......  (ii) (1 + x)–1 = 1 – x + x2 + x3 + .......  (iii) (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + .......  (iv) (1 + x)–2 = 1 – 2x + 3x2 – 4x3 + .......  6. 7. Exponential Series (a) ex = 1 + x x2 x3  1    .......... ; where x may be any real or complex number & e = Lim 1   n  1! 2! 3!  n (b) ax = 1 + x x2 x3 l na  l n 2 a  l n 3a + .......  , where a > 0 1! 2! 3! Logarithmic Series x2 x3 x4   (a) l n (1 + x) = x – + .....  , where – 1 < x  1 2 3 4 x 2 x3 x 4   (b) l n (1 – x) = – x – – .....  , where – 1  x < 1 2 3 4 (c) l n  1  x   x 3 x 5  2  x    ......  | x | 1 1  x   3 5  n