Question

The value of $4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right)}{2!}{{\left( \log 2 \right)}^{2}}+\dfrac{2\left( 1+{{2}^{3}} \right)}{3!}{{\left( \log 2 \right)}^{3}}+....\ is$
A. 10
B. 12
C. $\log \left( {{3}^{2}}{{.4}^{2}} \right)$
D. $\log \left( {{2}^{2}}{{.3}^{2}} \right)$

Answer 1

Hint: Expand the given entity. It gives the expansion form of ${{e}^{\log 2\ }}and\ {{e}^{2\log 2}}$. Substitute the exponent values and simplify the expression obtained until you get a whole number.

Complete step-by-step answer:
Given us $4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}+\dfrac{2\left( 1+{{2}^{3}} \right){{\left( \log 2 \right)}^{3}}}{3!}$
Take the first number 4 it can split as 2 + 2;
Take 2nd value, $2\left( 1+2 \right)\log 2$ it can be split us as;
$\left( 2+2\times 2 \right)\log 2=2\log 2+4\log 2$
Similarly,
$\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}=\dfrac{\left( 2+{{2}^{3}} \right){{\left( \log 2 \right)}^{2}}}{2!}=\dfrac{2{{\left( \log 2 \right)}^{2}}}{2!}+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}$
The other values can be similarly splitted,
$\begin{aligned} & 4+2\left( 1+2 \right)\log 2+\dfrac{2\left( 1+{{2}^{2}} \right){{\left( \log 2 \right)}^{2}}}{2!}+........ \\\ & =2+2+2\log 2+4\log 2+\dfrac{2{{(\log 2)}^{2}}}{2!}+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}+..... \\\ & =\left( 2+2\log 2+\dfrac{2{{\left( \log 2 \right)}^{2}}}{2!}+..... \right)+\left( 2+4\log 2+\dfrac{{{2}^{3}}{{\left( \log 2 \right)}^{2}}}{2!}+.... \right) \\\ \end{aligned}$
Taking 2 common from both brackets
$=2\left( 1+\log 2+\dfrac{{{\left( \log 2 \right)}^{2}}}{2!}+..... \right)+2\left( 1+2\log 2+\dfrac{{{\left( 2\log 2 \right)}^{2}}}{2!}+.... \right)$
This is the fourier expansion of general expansion of $_{e}{{\log }^{2}}$.
$\therefore {{\ }_{e}}{{\log }^{2}}=1+\log 2+\dfrac{{{\left( \log 2 \right)}^{2}}}{2!}+.....$
Similarly $1+2\log 2\dfrac{{{\left( 2\log 2 \right)}^{2}}}{2!}+....$ is the expansion of $_{e}{{\log }^{2\log 2}}$
$\therefore$ It can be written as,
$2\left( {{e}^{\log 2}} \right)+2\left( {{e}^{2\log 2}} \right)$
We know that ${{e}^{\log x}}=x$
$\begin{aligned} & \therefore \ {{e}^{\log 2}}=2 \\\ & {{e}^{2\log 2}}={{e}^{\log {{2}^{2}}}}={{2}^{2}}=4 \\\ & \Rightarrow 2\times 2+2\times 4=4+8=12 \\\ \end{aligned}$
Therefore, the correct answer is option B.

Note: Splitting of the entity is important in the beginning. So as to form the expansion form of ${{e}^{\log x}}$.