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Question

Find the sum of $$n$$ terms of the series whose $$nth$$ term is :
$$(2n-1)^{2}$$


Solution

$$\sum\ (2n-1)^2$$
$$\sum 4n^2-4n+1$$
$$4\sum n^2-4\sum\ n+\sum1$$
$$\dfrac{4n({n+1})(2n+1)}{6}+\dfrac{4n(n+1)}{2}+n$$


Mathematics

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