Question

Find the sum upto 15 term
$1 + 6 + \frac{9( {1}^{2} + {2}^{2} + {3}^{2}) }{7} + \frac{12( {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} ) }{9} + \frac{15( {1}^{2} + {2}^{2} +... + {5}^{2}) }{11} + ...$
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Answer 1

Pre-requisite Knowledge :-

• $\displaystyle\sum\limits_{n = 1}^kn = \dfrac{n(n+1)}{2}$

• $\displaystyle\sum\limits_{n = 1}^kn^2 = \dfrac{n(n+1)(2n+1)}{6}$

• $\displaystyle\sum\limits_{n = 1}^kn^3 = \left\{\dfrac{n(n+1)}{2}\right\}^2$

Now, let's solve the given problem!

We are given the series,

$\longrightarrow 1+6+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+\dfrac{15(1^2+2^2+...+5^2)}{11}+...\sf11\: terms$

Now, we have to form symmetry to each term ( or we have to make all the term of the series similar to each other )

$\longrightarrow \dfrac{3(1^2)}{3}+\dfrac{6(1^2+2^2)}{5}+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+\dfrac{15(1^2+2^2+...+5^2)}{11}+...\sf11\: terms$

Now we have to write general term of the given series. We can write the general term by noticing the pattern involved in the series.

Therefore,

$\longmapsto T_n = \dfrac{3n (1^2+2^2+3^2+...+n^2)}{(2n+1)}$

$\longmapsto T_n = \dfrac{3n\times\dfrac{n(n+1)(2n+1)}{6}}{(2n+1)}$

$\longmapsto T_n = \dfrac{3n\times n(n+1)(2n+1)}{6\times(2n+1)}$

$\underline{\boxed{ \longmapsto T_n = \dfrac{n^2(n+1)}{2}}}$

Therefore, this is the general term of the series.

Now, the sum of the series upto 15 terms is given by,

$\displaystyle \implies\sum\limits_{n=1}^{15}T_n$

$\displaystyle\implies\sum\limits_{n=1}^{15} \dfrac{n^3 + n^2}{2}$

$\displaystyle\implies\dfrac{1}{2}\left[\sum\limits_{n=1}^{15}n^3 + n^2\right]$

$\displaystyle\implies \dfrac{1}{2}\left[\sum\limits_{n=1}^{15}n^3 + \sum\limits_{n=1}^{15}n^2\right]$

${\displaystyle\implies \dfrac{1}{2}\left[\left(\dfrac{15(15+1)}{2}\right)^2 + \dfrac{15(15+1)(2(15)+1)}{6}\right]}$

${\displaystyle\implies \dfrac{1}{2}\left[\left(\dfrac{15\times 16}{2}\right)^2 + \dfrac{15\times 16\times (30+1)}{6}\right]}$

${\displaystyle\implies \dfrac{1}{2}\left[\left(\dfrac{15\times 8}{1}\right)^2 + \dfrac{5\times 8\times 31}{1}\right]}$

$\displaystyle\implies \dfrac{1}{2}\left[\left(120\right)^2 + 1240 \right]$

$\displaystyle\implies \dfrac{1}{2}\left[14400 + 1240 \right]$

$\displaystyle\implies \dfrac{1}{2}\left[15640 \right]$

$\displaystyle\implies 7820$

Therefore,

$\boxed{ 1+6+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+... = 7820 }$

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