24. 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.
Answer:
It is known that (k + 1)th term, (Tk+1), in the binomial expansion of (a + b)n is given by
Tk+1 = \(^nC_ra^{n-k}b^k\) .
Therefore, (r – 1)th term in the expansion of (x + 1)n is
(r + 1) term in the expansion of (x + 1)n is
rth term in the expansion of (x + 1)n is
Therefore, the coefficients of the (r – 1)th , rth, and (r + 1)th terms in the expansion of (x + 1)n \(^nC_{r-2},\,^nC_{r-1}\) and \(^nC_r\) are respectively. Since these coefficients are in the ratio 1:3:5, we obtain
Multiplying (1) by 3 and subtracting it from (2), we obtain
4r – 12 = 0
⇒ r = 3
Putting the value of r in (1), we obtain
n – 12 + 5 = 0
⇒ n = 7
Thus, n = 7 and r = 3
25. 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.
Answer:
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by
Tr+1 = \(^nC_ra^{n-r}b^r\) .
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n , we obtain
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
Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1 , we obtain
From (1) and (2), it is observed that
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.
26. Find a positive value of m for which the coefficient of x 2 in the expansion (1 + x) m is 6.
Answer:
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by
Tr+1 = \(^nC_ra^{n-r}b^r\) .
Assuming that x2 occurs in the (r + 1)th term of the expansion (1 +x)m, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain r = 2
Therefore, the coefficient of x2 is \(^mC_2\)
It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
Thus, the positive value of m, for which the coefficient of x2 in the expansion (1 + x)m is 6, is 4.
27. Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Answer:
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by
Tr+1 = \(^nC_ra^{n-r}b^r\) .
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
Dividing (2) by (1), we obtain
Dividing (3) by (2), we obtain
From (4) and (5), we obtain
Substituting n = 6 in equation (1), we obtain a6
= 729
⇒ a = \(\sqrt[6]{729}\) = 3
From (5), we obtain
Thus, a = 3, b = 5, and n = 6.
28. Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
Answer:
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by
Tr+1 = \(^nC_ra^{n-r}b^r\) .
Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9 , we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Thus, the coefficient of x2 is
Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9 , we obtain
Comparing the indices of x in x3 and in Tk+ 1, we obtain k = 3
Thus, the coefficient of x3 is
It is given that the coefficients of x2 and x3 are the same.
Thus, the required value of a is 9/7.
29. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
Answer:
Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7 , can be expanded as
The complete multiplication of the two brackets is not required to be carried out.
Only those terms, which involve x5 , are required.
The terms containing x5 are
Thus, the coefficient of x5 in the given product is 171.