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

$\begin{array}{1 1}\large\frac{n(4n^2-5)}{3} \\\large\frac{n(4n^2+1)}{3} \\\large\frac{n(4n^2-1)}{3} \\\large\frac{n(4n^2-n-1)}{3} \end{array}$

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A)
Toolbox:
• Sum of $n$ terms of any series $=S_n=\sum t_n$
• $\sum (A+B)=\sum A+\sum B$
• $\sum k.A=k.\sum A$ where $k$ is a constant.
• $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
• $\sum n=\large\frac{n(n+1)}{2}$
• $\sum 1=n$
Given $t_n=(2n-1)^2=4n^2-4n+1$
We know that sum of $n$ terms of the series $=S_n=\sum t_n=\sum(4n^2-4n+1)$
$\Rightarrow\:S_n=4\sum n^2-4.\sum n+\sum 1$
$\qquad\:=4.\large\frac{n(n+1)(2n+1)}{6}$$-4.\large\frac{n(n+1)}{2}$$+n$
$\qquad\:=n\bigg[\large\frac{2(n+1)(2n+1)}{3}$$-2(n+1)+1\bigg]$
$\qquad\:=n\bigg[\large\frac{4n^2+6n+2-6n-6+3}{3}\bigg]$
$\qquad\:=\large\frac{n(4n^2-1)}{3}$