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.
