Question

Coefficient if $x^{32}$ in the expansion of ${(x^4-\frac{1}{x^3})}^{15}$ is:

• A. $1156$
• B. $1365$
• C. $-965$
• D. none of the above

Answer 1

Answer: (B)

$1365$

Given

${(x^4-\frac{1}{x^3})}^{15}$

General term of given expansion is

$T_{r+1}={}^{15}{C}_r{x}^{60-4r}\frac{1}{x^{3r}}(-1{)}^r$

$T_{r+1}={}^{15}{C}_r{x}^{60-7r}(-1{)}^r-----(1)$

Comparing term having power $x^{32}$ with $T_{r+1}={}^{15}{C}_r{x}^{60-7r}(-1{)}^r$ we get

$60-7r=32\Rightarrow r=4$

Hence coefficient of term having power $x^{32}$ is ${}^{15}{C}_4(-1{)}^4$

${}^{15}{C}_4=\frac{15!}{4!11!}=\frac{15\times 14\times 13\times 12}{4\times 3\times 2}=1365$