Question
Find the sum of the series whose $n^{th}$ term is:$(2n - 1)^2$
✓
Answer
$T_n = (2n - 1)^2$
Let $S_n$ be the sum of $n$ terms of the given series.
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(2\text{k}-1)^2$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(4\text{k}^2+1-4\text{k})$
$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\sum\limits^{\text{n}}_{\text{k}=1}1-4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big[\frac{4(2\text{n}+1)}{3}-4\Big]+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{8\text{n}+4-12}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{8\text{n}-8}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=4\text{n}(\text{n}+1)\Big(\frac{\text{n}-1}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(4\text{n}^2+4\text{n}-4\text{n}-4+3)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(4\text{n}^2-1)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(2\text{n}-1)(2\text{n}+1)$
Let $S_n$ be the sum of $n$ terms of the given series.
Now,
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(2\text{k}-1)^2$
$\Rightarrow\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}(4\text{k}^2+1-4\text{k})$
$\Rightarrow\text{S}_\text{n}=4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\sum\limits^{\text{n}}_{\text{k}=1}1-4\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$\Rightarrow\text{S}_\text{n}=\frac{4\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}-\frac{4\text{n}(\text{n}+1)}{2}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big[\frac{4(2\text{n}+1)}{3}-4\Big]+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{8\text{n}+4-12}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big(\frac{8\text{n}-8}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=4\text{n}(\text{n}+1)\Big(\frac{\text{n}-1}{3}\Big)+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(4\text{n}^2+4\text{n}-4\text{n}-4+3)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(4\text{n}^2-1)$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}}{3}(2\text{n}-1)(2\text{n}+1)$
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