TOPICS
Unit-8(Examples)

Example-1 :-  Expand (x2 + 3/x)4, x ≠ 0

Solution :-
  (x2 + 3/x)4 
= 4 C0(x2)4 + 4C1(x2)3(3/x) + 4C2(x2)2(3/x)2 + 4C3(x2)(3/x)3 + 4C4(3/x)4
= x8 + 4x6 . 3/x + 6x4 . 9/x2 + 4x2 . 27/x3 + 81/x4
= x8 + 12x5 + 54x2 + 108/x + 81/x4
   

Example-2 :-  Compute (98)5.

Solution :-
  985 = (100 - 2)5 
= 5C0(100)5 - 5C1(100)4.2 + 5C2(100)3.22 - 5C3(100)2.23 + 5C4(100).24 - 5C5(2)5
= 100000000000 - 5 x 100000000 x 2 + 10 x 1000000 x 4 - 10 x 10000 x 8 + 5 x 100 x 16 - 32
= 10040008000 - 1000800032
= 9039207968
   

Example-3 :-  Which is larger (1.01)1000000 or 10,000?

Solution :-
  (1.01)1000000 = (1 + 0.01)1000000
= 1000000C0 + 1000000C1(0.01) + other positive terms 
= 1 + 1000000 × 0.01 + other positive terms
= 1 + 10000 + other positive terms
> 10000
  Hence (1.01)1000000 > 10000
    

Example-4 :-  Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25.

Solution :-
  For two numbers a and b if we can find numbers q and r such that a = bq + r, 
  then we say that b divides a with q as quotient and r as remainder. 
  Thus, in order to show that  6n – 5n leaves remainder 1 when divided by 25, 
  we prove that 6n – 5n = 25k + 1, where k is some natural number.
  We have (1 + a)n = nC0 + nC1a + nC2a2 + ... + nCnan 
  For a = 5, we get          
  (1 + 5)n = nC0 + nC15 + nC252 + ... + nCn5n i.e. 
  (6)n = 1 + 5n + 52.nC2 + 53.nC3 + ... + 5n i.e. 
  6n – 5n = 1 + 52(nC2 + nC35 + ... + 5n-2) or 
  6n – 5n = 1 + 25(nC2 + 5 .nC3 + ... + 5n-2) or 
  6n – 5n = 25k + 1     
  where k =  nC2 + 5 .nC3 + ... + 5n-2. 
  This shows that when divided by 25, 6n – 5n leaves remainder 1.
   

Example-5 :-  Find a if the 17th and 18th terms of the expansion (2 + a)50 are equal.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 =  nCr xn-r yr. 
  For the 17th term, we have, r + 1 = 17, i.e., r = 16 
  Therefore, T17 = T16+1 = 50C16 (2)50–16 a16 = 50C16 234 a16. 
  Similarly, T18 = 50C17 233 a17 
  Given that T17 = T18
  So, 50C16 (2)34 a16 = 50C17 (2)33 a17 
  Therefore, 
  [50C16 (2)34] ÷ [50C17 (2)33] = a17 ÷ a16
  a = 50!/(16!34!) x (17!33!)/50! x 2 = 1
   

Example-6 :-  Show that the middle term in the expansion of (1+x)2n is [1.3.5...(2n-1)]/n! 2n xn, where n is a positive integer.

Solution :-
  As 2n is even, the middle term of the expansion (1 + x)2n is (2n/2 + 1)th,
  i.e., (n + 1)th term which is given by,
  binomial theorem
    

Example-7 :-  Find the coefficient of x6y3 in the expansion of (x + 2y)9.

Solution :-
  Suppose x6y3 occurs in the (r + 1)th term of the expansion (x + 2y)9. 
  Now Tr+1  =  9Cr x9-r (2y)r  =  9Cr 2r . x9-r . yr . 
  Comparing the indices of x as well as y in x6y3 and in  Tr+1 , we get r = 3. 
  Thus, the coefficient of x6y3 is
  9C3 23 = 9!/(3!6!) x 23 = 672
   

Example-8 :-  The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n.

Solution :-
  Given that second term T2 = 240
  We have T2 = nC1 xn-1 . a 
  So nC1 xn-1 . a = 240 ... (1) 
  Similarly nC2 xn-2 a2 = 720 ... (2) 
  and nC3 xn-3 a3 = 1080 ... (3) 
  Dividing (2) by (1), we get
  [nC2 xn-2 a2] ÷ [nC1 xn-1 . a] = 720 ÷ 240
  [(n-1)!.a] ÷ [(n-2)!.x] = 6   ....(4)
  a/x = 6/(n-1)
  Dividing (3) by (2), we have
  [nC3 xn-3 a3] ÷ [nC2 xn-2 a2] = 1080 ÷ 720
  a/x = 9/[2(n-2)] ....(5)
  From (4) and (5),
  6/(n-1) = 9/[2(n-2)]
  9n - 9 = 12n - 24
  12n - 9n = 24 - 9
  3n = 15
   n = 5 
  Hence, from (1), 5x4a = 240, 
  and from (4), a/x = 3/2
  Solving these equations for a and x, we get x = 2 and a = 3. 
   

Example-9 :-  The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio1: 7 : 42. Find n.

Solution :-
  Suppose the three consecutive terms in the expansion of (1 + a)n are (r – 1)th, rth and (r + 1)th terms. 
  The (r – 1)th term is nCr-2 ar-2, and its  coefficient is nCr-2. 
  Similarly, the coefficients of rth and (r + 1)th terms  are nCr-1 and nCr, respectively. 
  Since the coefficients are in the ratio 1 : 7 : 42, so we have,
  nCr-2 : nCr-1 = 1/7,  i.e., n – 8r + 9 = 0 ... (1)
  nCr-1 : nCr = 7/42 , i.e., n – 7r + 1 = 0 ... (2)
  Solving equations(1) and (2), we get, n = 55.
    
CLASSES

Connect with us:

Copyright © 2015-17 by a1classes. All Rights Reserved.

www.000webhost.com