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168 . The sum of the series $(1+2)+\left(1+2+2^{2}\right)+$ $\left(1+2+2^{2}+2^{3}\right)+$ upto $n$ terms is
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Verified Answer
The correct answer is:
$2^{n+2}-n-4$
Here, $T_{n}=1+2+2^{2}+2^{3}+\cdots+2^{n}$
$$
\begin{aligned}
&=\frac{1\left(2^{n+1}-1\right)}{2-1}=2^{n+1}-1 \\
\therefore \Sigma T_{n}=& \Sigma\left(2^{n+1}-1\right)=\Sigma 2^{n+1}-\Sigma 1 \\
=&\left(2^{2}+2^{3}+\cdots+2^{n+1}\right)-n=2\left(2+2^{2}+\cdots+2^{n}\right)-n \\
=& \frac{4\left(2^{n}-1\right)}{2-1}-n=2^{n+2}-n-4
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{1\left(2^{n+1}-1\right)}{2-1}=2^{n+1}-1 \\
\therefore \Sigma T_{n}=& \Sigma\left(2^{n+1}-1\right)=\Sigma 2^{n+1}-\Sigma 1 \\
=&\left(2^{2}+2^{3}+\cdots+2^{n+1}\right)-n=2\left(2+2^{2}+\cdots+2^{n}\right)-n \\
=& \frac{4\left(2^{n}-1\right)}{2-1}-n=2^{n+2}-n-4
\end{aligned}
$$
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