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