Question

If $n$ is a positive integer and the coefficient of $x^{10}$ in the expansion of $(1+x{)}^{15}$ is equal to the coefficient of $x^5$ in the expansion of $(1-x{)}^{-n}$, then $n=$

• A. 15
• B. 12
• C. 11
• D. 10

Answer 1

Answer: (C)

In the expansion of $(1+x{)}^{15}$ the coefficient of
$x^{10}={}^{15}{C}_{10}$
And the expansion of
$(1-x{)}^{-n}=1+nx+\frac{n(n+1)}{2!}{x}^2+\frac{n(n+1)(n+2)x^3}{3!}+\dots$
So, coefficient of $x^5=\frac{n(n+1)(n+2)(n+3)(n+4)}{5!}$
According to the question,
$\frac{n(n+1)(n+2)(n+3)(n+4)}{5!}={}^{15}{C}_{10}$
$=\frac{15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$
$\Rightarrow \frac{n(n+1)(n+2)(n+3)(n+4)}{5!}=7\times 13\times 3\times 11$
$\Rightarrow n(n+1)(n+2)(n+3)(n+4)=5!(7\times 13\times 3\times 11)$
$\Rightarrow n(n+1)(n+2)(n+3)(n+4)=11\times 12\times 13\times 14\times 15$
By comparing, we get
$n=11$