Question

The tenth term in the expansion of ${(2x^2+\frac{1}{x})}^{12}$ is:

• A. $\frac{1760}{x^3}$
• B. $-\frac{1760}{x^3}$
• C. $\frac{1760}{x^2}$
• D. none of the above

Answer 1

Answer: (A)

$\frac{1760}{x^3}$

Given

${(2x^2+\frac{1}{x})}^{12}$

General term of given expansion is

$T_{r+1}={\sum }_{r=0}^{12}{}^{12}{C}_r(2x^2{)}^{12-r}{\left(\frac{1}{x}\right)}^r$

We need ${10}^{th}$ term hence put $r=9$

$T_{9+1}={}^{12}{C}_9(2x^2{)}^{12-9}{\left(\frac{1}{x}\right)}^9$

$T_{10}={}^{12}{C}_9(2x^2{)}^3{\left(\frac{1}{x}\right)}^9$

$T_{10}={}^{12}{C}_9\cdot 2^3(x{)}^{6-9}$

$T_{10}=\frac{12!}{9!3!}\cdot 8(x{)}^{-3}$

$T_{10}=\frac{12\times 11\times 10}{3\times 2\times x^3}\cdot 8$

$T_{10}=\frac{1760}{x^3}$