Question

The coefficient of $x^{39}$ in the expansion of ${(x^4-\frac{1}{x^3})}^{15}$ is:

• A. $455$
• B. $-455$
• C. $105$
• D. none of the above

Answer 1

The correct option is B $-455$

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 $x^{39}$ with $T_{r+1}={}^{15}{C}_r{x}^{60-7r}(-1{)}^r$ we get

$60-7r=39\Rightarrow r=3$

Hence coefficient of term having power of $x$ $x^{39}$ is ${}^{15}{C}_3(-1{)}^3$

$-{}^{15}{C}_3=-\frac{15!}{3!12!}=-\frac{15\times 14\times 13}{3\times 2}=-455$