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The sum of the first $n$ terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots \ldots \ldots$. is
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Verified Answer
The correct answer is:
$n+2^{-n}-1$
The sum of the first $n$ terms is
$S_n=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+\left(1-\frac{1}{2^4}\right)$ $+\ldots \ldots+\left(1-\frac{1}{2^n}\right)$
$=n-\left\{\frac{1}{2}+\frac{1}{2^2}+\ldots . .+\frac{1}{2^n}\right\}$
$=n-\frac{1}{2}\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)$$=n-\left(1-\frac{1}{2^n}\right)=n-1+2^{-n}$
Trick: Check for
$n=1,2$ i.e. $\quad S_1=\frac{1}{2}, S_2=\frac{5}{4}$
$\Rightarrow S_1=\frac{1}{2}$ and $S_2=2+2^{-2}-1=\frac{5}{4}$.
$S_n=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+\left(1-\frac{1}{2^4}\right)$ $+\ldots \ldots+\left(1-\frac{1}{2^n}\right)$
$=n-\left\{\frac{1}{2}+\frac{1}{2^2}+\ldots . .+\frac{1}{2^n}\right\}$
$=n-\frac{1}{2}\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)$$=n-\left(1-\frac{1}{2^n}\right)=n-1+2^{-n}$
Trick: Check for
$n=1,2$ i.e. $\quad S_1=\frac{1}{2}, S_2=\frac{5}{4}$
$\Rightarrow S_1=\frac{1}{2}$ and $S_2=2+2^{-2}-1=\frac{5}{4}$.
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