TOPICS
Miscellaneous

Example-1 :-  Find the term independent of x in the expansion of (3x2/2 - x/3)6.

Solution :-
  We have, Tr+1 
= 6Cr (3x2/2)6-r (-1/3x)r
= 6Cr (3/2)6-r (x2)6-r (-1)r (1/x)r (1/3)r
= (-1)r 6Cr [36-2r/26-r] x12-3r
  The term will be independent of x if the index of x is zero, i.e., 12 – 3r = 0. Thus, r = 4
  Hence 5th term is independent of x and is given by (-1)4 6C4 [36-8/26-4] x12-12 = 5/12
   

Example-2 :-  If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r + 1) + 4r2 – 2 = 0.

Solution :-
  The (r + 1)th term in the expansion is  nCrar. 
  Thus it can be seen that ar occurs in the (r + 1)th term, and its coefficient is nCr. 
  Hence the coefficients of ar-1, ar and ar+1 are nCr-1, nCr and nCr+1, respectively.  
  Since these coefficients are in arithmetic progression, so we have, nCr-1 + nCr+1 = 2.nCr. 
  This givesbinomial theorem
  r(r + 1) + (n – r) (n – r + 1) = 2 (r + 1) (n – r + 1) or 
  r2 + r + n2 – nr + n – nr + r2 – r = 2(nr – r2 + r + n – r + 1)
  n2 – 4nr – n + 4r2 – 2 = 0  
  n2 – n (4r + 1) + 4r2 – 2 = 0 
   

Example-3 :-  Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n-1.

Solution :-
  As 2n is even so the expansion (1 + x)2n has only one middle term which is (2n/2 + 1)th  i.e., (n + 1)th  term. 
  The (n + 1)th term is 2nCn xn. 
  The coefficient of xn is 2nCn 
  Similarly, (2n – 1) being odd, the other expansion has two middle terms, [(2n-1+1)/2]th and 
  [(2n-1+1)/2 + 1]th i.e., nth and (n + 1)th terms. 
  The coefficients of these terms are 2n-1Cn-1 and 2n-1Cn, respectively. 
  Now, 2n-1Cn-1 +  2n-1Cn = 2nCn      [As nCr-1 + nCr = n+1Cr]. as required.
    

Example-4 :-  Find the coefficient of a4 in the product (1 + 2a)4 (2 – a)5 using binomial theorem.

Solution :-
  We first expand each of the factors of the given product using Binomial Theorem. 
  We have (1 + 2a)4
= 4C0 + 4C1 (2a) + 4C2 (2a)2 + 4C3 (2a)3 + 4C4 (2a)4
= 1 + 4 (2a) + 6(4a2) + 4 (8a3) + 16a4. 
= 1 + 8a + 24a2 + 32a3 + 16a4

  Now, (2 – a)5
= 5C0 (2)55C1 (2)4 (a) + 5C2 (2)3 (a)25C3 (2)2 (a)3 + 5C4 (2) (a)45C5 (a)5 
= 32 – 80a + 80a2 – 40a3 + 10a4 – a5

  Thus (1 + 2a)4 (2 – a)5 = (1 + 8a + 24a2 + 32a3 + 16a4) (32 – 80a + 80a2 – 40a3 + 10a4 – a5) 
  The complete multiplication of the two brackets need not be carried out. 
  We write only those terms which involve a4. 
  This can be done if we note that ar. a4-r = a4. 
  The terms containing a4 are 1 (10a4) + (8a) (–40a3) + (24a2) (80a2) + (32a3) (– 80a) + (16a4) (32) = – 438a4
  Thus, the coefficient of a4 in the given product is – 438. 
   

Example-5 :-  Find the rth term from the end in the expansion of (x + a)n.

Solution :-
  There are (n + 1) terms in the expansion of (x + a)n. 
  Observing the terms we can say that the first term from the end is the last term, i.e., 
  (n + 1)th term of the expansion and n + 1 = (n + 1) – (1 – 1). 
  The second term from the end is the nth term of the expansion, and  n = (n + 1) – (2 – 1). 
  The third  term from the end is the (n – 1)th term of the expansion and n – 1 = (n + 1) – (3 – 1) and so on. 
  Thus rth term from the end will be term number (n + 1) – (r – 1) = (n – r + 2) of the expansion. 
  And the (n – r + 2)th term is nCn–r+1 xr-1 an-r+1.
   

Example-6 :-  Find the term independent of x in the expansion of binomial theorem

Solution :-
  binomial theorem
  Since we have to find a term independent of x, i.e., term not having x, so take (18 - 2r)/3 = 0
  We get r = 9. 
  The required term is 18C9 1/29
   

Example-7 :-  The sum of the coefficients of the first three terms in the expansion of (x - 3/x2)m , x ≠ 0, m being a natural number, is 559. Find the term of the expansion containing x3.

Solution :-
  The coefficients of the first three terms of (x - 3/x2)m are mC0, (–3) mC1 and 9 mC2. 
  Therefore, by the given condition, we have mC0 (–3) mC1 + 9 mC2 = 559, i.e., 1 – 3m + [9m(m-1)/2] = 559
  which gives  m = 12 (m being a natural number).
  Now, Tr+1 =  12Cr x12-r (-3/x2)r =  12Cr (–3)r . x12-3r
  Since we need the term containing x3, so put 12 – 3r = 3 i.e., r = 3. 
  Thus, the required term is 12C3 (–3)3  x3,  i.e., – 5940 x3. 
   

Example-8 :-  If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal, find r.

Solution :-
  The coefficients of (r – 5)th and (2r – 1)th terms of the expansion (1 + x)34 are 34Cr-6 and 34C2r-2, respectively. 
  Since they are equal so, 34Cr-6 = 34C2r-2
  Therefore, either r – 6 = 2r – 2  or  r – 6 = 34 – (2r – 2) 
  [Using the fact that if nCr = nCp, then either r = p or r = n – p] 
  So, we get r = – 4 or r = 14. 
  r being a natural number, r = – 4 is not possible. 
  So, r = 14.
   
CLASSES

Connect with us:

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

www.000webhost.com