# Answer The Following Question With Explanation ?

## $\left({1}^{2}\right) + \left({1}^{2} + {2}^{2}\right) + \left({1}^{2} + {2}^{2} + {3}^{2}\right) + \ldots \ldots \ldots \ldots$ Summation Of ${n}^{t h}$ term

Aug 1, 2018

$\frac{n}{12} {\left(n + 1\right)}^{2} \left(n + 2\right)$.

#### Explanation:

Observe that, the ${m}^{t h}$ term ${t}_{m}$ is given by,

${t}_{m} = {1}^{2} + {2}^{2} + \ldots + {m}^{2} = \frac{m}{6} \left(m + 1\right) \left(2 m + 1\right)$.

$\therefore \text{The Reqd. Sum } {s}_{n} = {\sum}_{m = 1}^{m = n} {t}_{m}$,

$= \sum \left\{\frac{m}{6} \left(m + 1\right) \left(2 m + 1\right)\right\}$,

$= \frac{1}{6} \sum \left\{2 {m}^{3} + 3 {m}^{2} + m\right\}$,

$= \frac{1}{6} \left\{2 \sum {m}^{3} + 3 \sum {m}^{2} + \sum m\right\}$,

$= \frac{2}{6} \cdot {n}^{2} / 4 {\left(n + 1\right)}^{2} + \frac{3}{6} \cdot \frac{n}{6} \left(n + 1\right) \left(2 n + 1\right) + \frac{1}{6} \cdot \frac{n}{2} \left(n + 1\right)$,

$= {n}^{2} / 12 {\left(n + 1\right)}^{2} + \frac{n}{12} \left(n + 1\right) \left(2 n + 1\right) + \frac{n}{12} \left(n + 1\right)$,

$= \frac{n}{12} \left(n + 1\right) \left[\left\{{n}^{2} + n\right\} + \left\{2 n + 1\right\} + 1\right]$,

$= \frac{n}{12} \left(n + 1\right) \left\{\left({n}^{2} + 3 n + 2\right)\right\}$,

$= \frac{n}{12} \left(n + 1\right) \left\{\left(n + 1\right) \left(n + 2\right)\right\}$.

$\Rightarrow \text{The Reqd. Sum} = \frac{n}{12} {\left(n + 1\right)}^{2} \left(n + 2\right)$.

Aug 1, 2018

${S}_{n} = \frac{n}{12} {\left(n + 1\right)}^{2} \left(n + 2\right)$

#### Explanation:

Let ,

$S$=${1}^{2} + \left({1}^{2} + {2}^{2}\right) + \left({1}^{2} + {2}^{2} + {3}^{2}\right) + \ldots + \left({1}^{2} + {2}^{2} + \ldots + {n}^{2}\right)$

$\therefore {t}_{n} = {1}^{2} + {2}^{2} + {3}^{2} + \ldots + {n}^{2}$

$\therefore {t}_{n} = {\sum}_{r = 1}^{n} {r}^{2} = \frac{n}{6} \left(n + 1\right) \left(2 n + 1\right)$

$\therefore {t}_{n} = \frac{1}{6} \left\{n \left(2 {n}^{2} + 2 n + n + 1\right)\right\} = \frac{1}{6} \left\{2 {n}^{3} + 2 {n}^{2} + {n}^{2} + n\right\}$

$\therefore {t}_{n} = \frac{1}{6} \left\{2 {n}^{3} + 3 {n}^{2} + n\right\}$

$\therefore {S}_{n} = \frac{1}{6} \left\{2 {\sum}_{r = 1}^{n} {r}^{3} + 3 {\sum}_{r = 1}^{n} {r}^{2} + {\sum}_{r = 1}^{n} r\right\}$

$= \frac{1}{6} \left\{\cancel{2} {n}^{2} / {\cancel{4}}^{2} {\left(n + 1\right)}^{2} + \cancel{3} \frac{n}{\cancel{6}} ^ 2 \left(n + 1\right) \left(2 n + 1\right) + \frac{n}{2} \left(n + 1\right)\right\}$

$= \frac{1}{6} \frac{n}{2} \left(n + 1\right) \left\{n \left(n + 1\right) + 2 n + 1 + 1\right\}$

$= \frac{n}{12} \left(n + 1\right) \left\{{n}^{2} + n + 2 n + 2\right\}$

$= \frac{n}{12} \left(n + 1\right) \left({n}^{2} + 3 n + 2\right)$

$\therefore {S}_{n} = \frac{n}{12} \left(n + 1\right) \left(n + 1\right) \left(n + 2\right)$

$\therefore {S}_{n} = \frac{n}{12} {\left(n + 1\right)}^{2} \left(n + 2\right)$
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Note:
$\left(1\right) {\sum}_{r = 1}^{n} r = \frac{n}{2} \left(n + 1\right)$
$\left(2\right) {\sum}_{r = 1}^{n} {r}^{2} = \frac{n}{6} \left(n + 1\right) \left(2 n + 1\right)$
$\left(3\right) {\sum}_{r = 1}^{n} {r}^{3} = {n}^{2} / 4 {\left(n + 1\right)}^{2}$