# 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. $\log \left( {{2}^{2}}{{.3}^{2}} \right)$

Last updated date: 22nd Mar 2023

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Answer

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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{align}

& 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{align}$

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{align}

& \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{align}$

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 ${{e}^{\log x}}$.

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{align}

& 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{align}$

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{align}

& \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{align}$

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 ${{e}^{\log x}}$.

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