Question

Find the sum of the series upto n terms whose nth term is given by,
2n² - 3n + 5​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm: \longmapsto T_{n} = 2 {n}^{2} - 3n + 5$

We have to find out the sum of the series.

$\displaystyle\rm: \longmapsto S _{n} = \sum^{n}_{k = 1} T_{k}$

$\displaystyle\rm: \longmapsto S _{n} = \sum^{n}_{k = 1} \big(2 {k}^{2} - 3k + 5 \big)$

$\displaystyle\rm: \longmapsto S _{n} = \sum^{n}_{k = 1}2 {k}^{2} - \sum^{n}_{k = 1} 3k + \sum^{n}_{k = 1}5$

$\displaystyle\rm: \longmapsto S _{n} = 2\sum^{n}_{k = 1}{k}^{2} - 3\sum^{n}_{k = 1}k +5n$

We know that:

$\displaystyle\rm: \longmapsto \sum^{n}_{k = 1}{k}^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$

$\displaystyle\rm: \longmapsto \sum^{n}_{k = 1}{k} = \dfrac{n(n + 1)}{2}$

Therefore:

$\displaystyle\rm: \longmapsto S _{n} = 2 \cdot \dfrac{n(n + 1)(2n + 1)}{6} - 3\cdot \dfrac{n(n + 1)}{2} +5n$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{2n(n + 1)(2n + 1)}{6} - \dfrac{9n(n + 1)}{6} + \frac{30n}{6}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{2n(n + 1)(2n + 1) - 9n(n + 1) + 30n}{6}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n(n + 1) \{2(2n + 1) - 9 \}+ 30n}{6}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n(n + 1) \{4n +2 - 9 \}+ 30n}{6}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n(n + 1)(4n - 7 )+ 30n}{6}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n}{6} \{ (n + 1)(4n - 7 )+ 30 \}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n}{6} \{4 {n}^{2} - 7n + 4n - 7 + 30 \}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n}{6} \{4 {n}^{2} - 3n-23\}$

$\displaystyle\rm: \longmapsto S _{n} = \dfrac{n}{6}(4 {n}^{2} - 3n-23)$

Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

$\displaystyle1. \: \: \rm \sum^{n}_{i = 1}k = kn$

$\displaystyle2. \: \: \rm \sum^{n}_{i = 1}i = \dfrac{n(n + 1)}{2}$

$\displaystyle3. \: \: \rm \sum^{n}_{i = 1} {i}^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$

$\displaystyle4. \: \: \rm \sum^{n}_{i = 1} {i}^{3} = \bigg( \dfrac{n(n + 1)}{2} \bigg)^{2}$