Question

Let the coefficient of ${10}^{th}$ term of an expansion be $a$ and $b$ be the power.
Expansion:${(2x^2+\frac{1}{x})}^{12}$
Find $a\times b$.

• A. $-5280$
• B. $4200$
• C. $-3460$
• D. $7360$

Answer 1

The correct option is A $-5280$

$(r+1{)}^{th}$ term is $t_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$
So, for ${10}^{th}$ term, $r=9$.
$\therefore t_{10}=t_{9+1}{=}^{12}{C}_9\left(2x^2{)}^{12-9}\right(\frac{1}{x}{)}^9$
$\therefore t_{10}=\frac{1760}{x^3}$
So, $a=1760,b=3$
Ans-Option $A$.