Question

Find the sum of below series upto n terms

1.2² + 2.3² + 3.4² + . . . ​

Answer 1

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

The nth term of the series is given as:

$\rm: \longmapsto T_{n} = n {(n + 1)}^{2}$

$\rm: \longmapsto T_{n} = n( {n}^{2} + 2n + 1)$

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

Therefore, the sum of the series will be:

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} T_{k}$

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} \big( {k}^{3} + 2 {k}^{2} + k \big)$

Separating the terms, we get:

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} {k}^{3} + \sum_{k = 1}^{n}2 {k}^{2} + \sum_{k = 1}^{n} k$

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} {k}^{3} + 2\sum_{k = 1}^{n}{k}^{2} + \sum_{k = 1}^{n} k$

We know that:

$\displaystyle\rm: \longmapsto \sum_{k = 1}^{n} {k}^{3} = \bigg[\frac{n(n + 1)}{2} \bigg]^{2}$

$\displaystyle\rm: \longmapsto \sum_{k = 1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{2}$

$\displaystyle\rm: \longmapsto \sum_{k = 1}^{n} {k}^{} = \frac{n(n + 1)}{2}$

Substituting the values, we get:

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} + 2 \cdot\frac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n +1 )}{2}$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} + \frac{n(n + 1)(2n + 1)}{3} + \dfrac{n(n +1 )}{2}$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{ {n}^{} {(n + 1)}^{} }{2} + \frac{2(2n + 1)}{3} + 1 \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{} {(n + 1)}^{} + 4(2n + 1) + 6 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{2} + 3n+ 8n +4 + 6 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{2} + 11n +10 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{2} + 6n + 5n+10 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3n(n +2) + 5(n+2)}{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{(3n + 5)(n+2)}{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)(n + 2)(3n + 5)}{12}$

★ Which is our required answer.

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

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

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

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

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

$\displaystyle5. \: \: \rm \sum^{n}_{i = 1} 2i = n(n+1)$

$\displaystyle6. \: \: \rm \sum^{n}_{i = 1} \big(2i-1\big)= n^{2}$

Answer 2

Answer:

$[tex]\textsf{\large{\underline{Solution}:}}$

The nth term of the series is given as:

$\rm: \longmapsto T_{n} = n {(n + 1)}^{2}$

$\rm: \longmapsto T_{n} = n( {n}^{2} + 2n + 1)$

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

Therefore, the sum of the series will be:

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} T_{k}$

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} \big( {k}^{3} + 2 {k}^{2} + k \big)$

Separating the terms, we get:

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} {k}^{3} + \sum_{k = 1}^{n}2 {k}^{2} + \sum_{k = 1}^{n} k$

$\displaystyle\rm: \longmapsto S_{n} = \sum_{k = 1}^{n} {k}^{3} + 2\sum_{k = 1}^{n}{k}^{2} + \sum_{k = 1}^{n} k$

We know that:

$\displaystyle\rm: \longmapsto \sum_{k = 1}^{n} {k}^{3} = \bigg[\frac{n(n + 1)}{2} \bigg]^{2}$

$\displaystyle\rm: \longmapsto \sum_{k = 1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{2}$

$\displaystyle\rm: \longmapsto \sum_{k = 1}^{n} {k}^{} = \frac{n(n + 1)}{2}$

Substituting the values, we get:

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} + 2 \cdot\frac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n +1 )}{2}$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} + \frac{n(n + 1)(2n + 1)}{3} + \dfrac{n(n +1 )}{2}$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{ {n}^{} {(n + 1)}^{} }{2} + \frac{2(2n + 1)}{3} + 1 \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{} {(n + 1)}^{} + 4(2n + 1) + 6 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{2} + 3n+ 8n +4 + 6 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{2} + 11n +10 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3 {n}^{2} + 6n + 5n+10 }{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{3n(n +2) + 5(n+2)}{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)}{2} \bigg[ \dfrac{(3n + 5)(n+2)}{6} \bigg]$

$\displaystyle\rm: \longmapsto S_{n} = \dfrac{n(n + 1)(n + 2)(3n + 5)}{12}$

★ Which is our required answer.

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

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

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

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

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

$\displaystyle5. \: \: \rm \sum^{n}_{i = 1} 2i = n(n+1)$

$\displaystyle6. \: \: \rm \sum^{n}_{i = 1} \big(2i-1\big)= n^{2}$[/tex]