Question

Find the coefficient of $a^5{b}^7$ in $(a-2b{)}^{12}$

• A. 12356
• B. -101356
• C. 101376
• D. -101376

Answer 1

The correct option is D -101376

It is know that $(r+1{)}^{th}$ term $\left(T_{r+1}\right)$ in the binomial expansion
of $(a+b{)}^n$ is given by $T_{r+1}{=}^n{C}_r{a}^{a-r}{b}^r$
Assuming that $a^5{b}^7$ occurs in the $(r+1{)}^{th}$ term of the expansion $(a-2b{)}^{12}$ we obtain
$T_{r+1}{=}^{12}{C}_r\left(a{)}^{12-r}\right(-2b{)}^r{=}^{12}{C}_r(-2{)}^r(a{)}^{12-r}(b{)}^r$
Comparing the indices of a and b in $a^5{b}^7$ and in $T_{r+1}$, we obtain $r=7$
Thus the coefficient of $a^5{b}^7$ is
${=}^{12}{C}_7(-2{)}^7=\frac{12!}{7!5!}\times 2^7=\frac{\mathrm{12.11.10.9.8.7}!}{\mathrm{5.4.3.2.7}!}{.2}^7=-(792\left)\right(128)=-101376$