Question

Find the ${15}^{th}$ term from the end in the expansion of $(4x+\frac{1}{16x^2}{)}^{21}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$p^{th}$ term from end is $(n+2-p{)}^{th}$ term from the beginning in the expansion of $(a+b{)}^n$
$\Rightarrow {15}^{th}$ term from the end is $(21+2-15{)}^{th}$ term from the beginning.
=$8^{th}$ term from the beginning
$\Rightarrow T_8={}^{21}{C}_7\left(4x{)}^{21-7}\right(\frac{1}{16x^2}{)}^7$
=${}^{21}{C}_7\left(4x{)}^{14}\right(\frac{1}{16x^2}{)}^7$
=${}^{21}{C}_7\left(4^{14}{x}^{14}\right)\times {16}^{-7}{x}^{-14}$
=${}^{21}{C}_7\left(4^{14}{x}^{14}\right)\times 4^{-14}{x}^{-14}$
=${}^{21}{C}_7$