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
$(1-x)^{-n}=1+n x+\frac{n(n+1)}{2 !} x^{2}+\frac{n(n+1)(n+2) x^{3}}{3 !}+\ldots$
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
$n= 11$