Question

The value of $4+2(1+2)\log 2+\frac{2(1+2^2)}{2!}{\left(\log 2\right)}^2+\frac{2(1+2^3)}{3!}{\left(\log 2\right)}^3+....$ is

• A. $10$
• B. $12$
• C. $\log (3^2.4^2)$
• D. $\log (2^2.3^2)$

Answer 1

The correct option is B $12$

$4+2(1+2)\log 2+\frac{1(1+2^2){\left(\log 2\right)}^2}{2!}+\frac{1(1+2^3){\left(\log 2\right)}^3}{3!}+...$
$=2\left(e^{\log 2}\right)+2\left(e^{2\log 2}\right)=2\times 2+2e^{\log 4}=12$