Question

find the sum of the series 3*1²+5*2²+7*3²+...... n terms​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

ꜰɪɴᴅ ᴛʜᴇ ꜱᴜᴍ ᴛᴏ ɴ ᴛᴇʀᴍꜱ ᴏꜰ ᴛʜᴇ ꜱᴇʀɪᴇꜱ 3 × 1² + 5 × 2² + 7 × 3² + …

♣ ᴀɴꜱᴡᴇʀ :

$\sf{The\:given\:series\:is\: 3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}+\ldots}$

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

$\bf{\therefore S_{n}=& \sum_{k=1}^{n} a_{k} }$

$\sf{=& \sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)=2 \sum_{k=1}^{n} }$

$\sf{k^{3}+\sum_{k=1}^{n} k^{2}}$

$\sf{=& 2\left[\dfrac{n(n+1)}{2}\right]^{2}+\dfrac{n(n+1)(2 n+1)}{6} }$

$\sf{=& \dfrac{n^{2}(n+1)^{2}}{2}+\dfrac{n(n+1)(2 n+1)}{6} }$

$\sf{=& \dfrac{n(n+1)}{2}\left[n(n+1)+\dfrac{2 n+1}{3}\right] }$

$\sf{=& \dfrac{n(n+1)}{2}\left[\dfrac{3 n^{2}+3 n+2 n+1}{3}\right] }$

$\sf{=& \dfrac{n(n+1)}{2}\left[\dfrac{3 n^{2}+5 n+1}{3}\right] }$

$\bf{=\dfrac{n(n+1)\left(3 n^{2}+5 n+1\right)}{6}}$