Question

find the sum of n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + .....​

Answer 1

$\Huge\bigstar\:\:\tt\underline\red{SOLUTION}\:\:\bigstar$

$\Large\sf \purple{The \:given \:series\: is\::}$

$\sf\purple{ 1^{2} \:+ \:(1^{2}\: +\: 2^{2}) \:+\: (1^{2} \: + \: 2^{2} \: + \: 3^{3}) \:+\: ...}$

$\sf\green{a_{n} \:=\:(1^{2}\:+\:2^{2}\:+\:3^{3}\:+\:.....\:+\:n^{2})}$

$\sf\red{\longrightarrow\:\:{\Large{\frac{n(n+1)(2n+1)}{6}}}}$

$\sf\red{\longrightarrow\:\:{\Large{\frac{n(2n^{2}+3n+1)}{6}\:=\:\frac{2^{3}+3n^{2}+n}{6}}}}$

$\sf\red{\longrightarrow\:\:{\Large{\frac{1}{3}{\small n^{3}}\:+\:\frac{1}{2}{\small n^{2}}\:+\:\frac{1}{6}{\small n}}}}$

$\large\bold\blue{\therefore\:\:}$$\large\sf\blue{\:S_{n}\:=\sum\limits_{k=1}^{n} a_{k}}$

$\large\sf\pink{\longrightarrow\:\:\sum\limits_{k=1}^{n}(\frac{1}{3}{\small n^{3}}\:+\:\frac{1}{2}{\small n^{2}}\:+\:\frac{1}{6}{\small n})}$

$\large\sf\pink{\longrightarrow\:\:\frac{1}{3}\sum\limits_{k=1}^{n}k^{3}\:+\:\frac{1}{2}\sum\limits_{k=1}^{n}k^{2}\:+ \:\frac{1}{6}\sum\limits_{k=1}^{n}k}$

$\large\sf\pink{\longrightarrow\:\:}$$\small\sf\pink{\frac{1}{3}\frac{n^{2}(n+1)^{2}}{(2)^{2}}+\frac{1}{2}×\frac{n(n+1)(2n+1)}{6}+\frac{1}{6}×\frac{n(n+1)}{2} }$

$\Large\sf\pink{\longrightarrow\:\:}$$\large\sf\pink{\frac{n(n+1)}{6}{\huge[}\frac{n(n+1)}{2}+\frac{(2n+1)}{2}+\frac{1}{2} {\huge]} }$

$\Large\sf\pink{\longrightarrow\:\:}$$\large\sf\pink{\frac{n(n+1)}{6}{\huge[}\frac{n^{2}+n+2n+2}{2} {\huge]} }$

$\Large\sf\pink{\longrightarrow\:\:}$$\large\sf\pink{\frac{n(n+1)}{6}{\huge[} \frac{n(n+1)+2(n+1)}{2}{\huge]} }$

$\Large\sf\pink{\longrightarrow\:\:}$$\large\sf\pink{\frac{n(n+1)}{6}{\huge[}\frac{(n+1)(n+2)}{2} {\huge]} }$

$\Large\sf\orange{\longrightarrow\:\:}$$\huge\sf\orange{\frac{n(n+1)^{2}(n+2)}{12}}$

Answer 2

Step-by-step explanation:

Steps are Given Above . .