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If the $n$ th term of an arithmetic progression is $2 n-1$, then what is the sum upto $n$ terms?
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The correct answer is:
$n^{2}$
Given $a_{n}=2 n-1$
$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}(2 n-1)$
$=2 \sum_{k=1}^{n} n-n=2 \cdot \frac{n(n+1)}{2}-n=n^{2}+n-n=n^{2}$
$\therefore S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}(2 n-1)$
$=2 \sum_{k=1}^{n} n-n=2 \cdot \frac{n(n+1)}{2}-n=n^{2}+n-n=n^{2}$
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