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Exercise - 8.2

Question-1 :-  Find the coefficient of x5 in (x + 3)8 .

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, 
  we obtain Tr+1 = 8Cr x8-r 3r
  Comparing the indices of x in x5 and in Tr+1, we obtain r = 3
  Thus, the coefficient of x5 is 8C3 33 = 8!/3!5! x 33 = 1512
   

Question-2 :- Find the coefficient of a5b7 in (a – 2b)12.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, 
  we obtain Tr+1 = 12Cr a12-r (-2b)r = 12Cr (-2)r a12-r (b)r
  Comparing the indices of a and b in a5 b7 and in Tr+1, we obtain r = 7
  Thus, the coefficient of a5b7 is 12C7 (-2)7 = 12!/7!5! x (-2)7 = (-792)(128) = -101376
   

Question-3 :-  Write the general term in the expansion of (x2 – y)6.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Thus, the general term in the expansion of (x2 – y)6 is 
  Tr+1 = 6Cr (x2)6-r (-y)r = 6Cr (-1)r x12-2r yr
   

Question-4 :-  Write the general term in the expansion of (x2 – yx)12, x ≠ 0.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Thus, the general term in the expansion of (x2 – yx)12 is 
  Tr+1 = 12Cr (x2)12-r (-yx)r 
      = 12Cr (-1)r x24-2r xr yr 
      = 12Cr (-1)r x24-r yr
   

Question-5 :-  Find the 4th term in the expansion of (x – 2y)12.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Thus, the 4th term in the expansion of (x – 2y)12 is 
  T4 = T3+1 
     = 12C3 x12-3 (-2y)3 
     = (-1)3 12!/3!9! x9 8y3 
     = -1760 x9 y3
   

Question-6 :-  Find the 13th term in the expansion of (9x - 1/3√x)18, x ≠ 0.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Thus, the 13th term in the expansion of (9x – 1/3√x)18 is 
  T13 = T12+1 
     = 18C12 (9x)18-12 (-1/3√x)12 
     = (-1)12 18!/12!6! (9)6 (x)6 (1/3)12 (1/√x)12
     = 18!/12!6! (3)12 (x)6 (1/3)12 (1/x)6
     = 18!/12!6!
     = 18564
   

Question-7 :-  Find the middle terms in the expansions of (3 - x3/6)7.

Solution :-
  It is known that in the expansion of (a + b)n, if n is odd, 
  then there are two middle terms, namely, [(n+1)/2]th term and [(n+1)/2 + 1] term.
  Therefore, the middle terms in the expansion of (3 - x3/6)7 are [(7+1)/2]th term and [(7+1)/2 + 1]th term.
  Now, T4 = T3+1 
       = 7C3 37-3 (-x3/6)3 
       = (-1)3 7!/3!4! 34 (x9/63)
       = -105x9/8
  Again, T5 = T4+1 
       = 7C4 37-4 (-x3/6)4 
       = (-1)4 7!/3!4! 33 (x12/64)
       = 35x12/48

  Thus, the middle terms in the expansion of (3 - x3/6)7 are -105x9/8 and 35x12/48.
   

Question-8 :-  Find the middle terms in the expansions of (x/3 + 9y)10.

Solution :-
  It is known that in the expansion of (a + b)n, if n is even, 
  then there are two middle terms, namely, [n/2 + 1]th term.
  Therefore, the middle terms in the expansion of (x/3 + 9y)10 are [10/2 + 1]th = 6th term.
  Now, T6 = T5+1 
       = 10C5 (x/3)10-5 (9y)5 
       = 95 10!/5!5! x5 (1/3)5 y5
       = 310 10!/5!5! x5 1/35 y5
       = 61236 x5 y5

  Thus, the middle terms in the expansion of (x/3 + 9y)10 is 61236 x5 y5.
   

Question-9 :-  In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m+n, 
  we obtain Tr+1 = m+nCr 1m+n-r ar = m+nCr ar
  Comparing the indices of a in am and in Tr+1, we obtain r = m
  Therefore, the coefficient of am is
  m+nCm = (m+n)!/[m!(m+n-m)!] = (m+n)!/m!n! .....(i)

  Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, 
  we obtain Tk+1 = m+nCk 1m+n-k ak = m+nCk ak
  Comparing the indices of a in an and in Tk+1, we obtain k = n
  Therefore, the coefficient of an is
  m+nCn = (m+n)!/[n!(m+n-n)!] = (m+n)!/m!n! .....(ii)

  Thus, from (i) and (ii), it can be observed that the coefficients of am and an in the expansion of (1 + a)m+n are equal.
   

Question-10 :-  The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Therefore, (r – 1)th term in the expansion of (x + 1)n is Tr-1 = nCr-2 xn-(r-2) 1r-2 = nCr-2 xn-r+2
  Now, rth term in the expansion of (x + 1)n is Tr = nCr-1 xn-(r-1) 1r-1 = nCr-1 xn-r+1
  Again, (r + 1)th term in the expansion of (x + 1)n is Tr+1 = nCr xn-r 1r

  Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)n are nCr-2, nCr-1 and nCr respectively. 
  Since these coefficients are in the ratio 1:3:5, we obtain
  nCr-2 : nCr-1 = 1 : 3
  n!/[(r-2)!(n-r+2)!] x [(r-1)!(n-r+1)!]/n! = 1/3
  (r-1)/(n-r+2) = 1/3
  n - r + 2 = 3r - 3
  n - 4r + 5 = 0 .....(i)

  nCr-1 : nCr = 3 : 5
  n!/[(r-1)!(n-r+1)!] x [r!(n-r)!]/n! = 3/5
  r/(n-r+1) = 3/5
  3n - 3r + 3 = 5r
  3n - 8r + 3 = 0 .....(ii)

  Multiplying (i) by 3 and subtracting it from (ii), we obtain 4r – 12 = 0, r = 3
  Putting the value of r in (i), we obtain n – 12 + 5 = 0, n = 7
  Thus, n = 7 and r = 3
   

Question-11 :-  Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n-1.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain Tr+1 = 2nCr (1)2n-r xr = 2nCr xr
  Comparing the indices of x in xn and in Tr+1, we obtain r = n
  Therefore, the coefficient of xn in the expansion of (1 + x)2n is 2nCn = (2n)!/n!n! = (2n)!/(n!)2  .......(i)

  Assuming that xn occurs in the (k + 1)th term of the expansion (1 + x)2n-1, we obtain Tk+1 = 2n-1Ck (1)2n-1-k xk = 2n-1Ck xk
  Comparing the indices of x in xn and Tk+1, we obtain k = n
  Therefore, the coefficient of xn in the expansion of (1 + x)2n-1 is 
  2n-1Ck = (2n-1)!/[n!(n-1)!] = (2n)!/(2.n!n!) = 1/2 x [(2n)!/(n!)2] .....(ii)

  From (i) and (ii), it is observed that
  1/2 x 2nCn = 2n-1Cn
  2nCn = 2.2n-1Cn
  Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n-1.
  Hence, proved.
   

Question-12 :-  Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Solution :-
  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. 
  Assuming that x2 occurs in the (r + 1)th term of the expansion (1 + x)m, we obtain Tr+1 = mCr (1)m-r xr = mCr xr
  Comparing the indices of x in x2 and in Tr+1, we obtain r = 2
  Therefore, the coefficient of x2 is mC2.
  It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
  Therefore, mC2 = 6
  m!/[2!(m-2)!] = 6
  [m x (m-1) x (m-2)!]/[2 x 1 x (m-2)!] = 6
  [m(m-1)]/2 = 6
  m2 - m = 12
  m2 - m - 12 = 0
  m2 - 4m + 3m - 12 = 0
  m(m-4) + 3(m-4) = 0
  (m + 3)(m - 4) = 0
  m = -3 or m = 4
  Thus, the positive value of m, for which the coefficient of x2 in the expansion (1 + x)m is 6, is 4.
   
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