Question

3²+7²+11²+...+ n terms

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Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given series is

$\rm :\longmapsto\: {3}^{2} + {7}^{2} + {11}^{2} + - - - n \: terms$

Now, 3, 7, 11, forms an AP series, with first term 3 and common difference 4. So nth term of series is

$\rm :\longmapsto\:T_n = {\bigg[3 + (n - 1)4\bigg]}^{2}$

$\rm :\longmapsto\:T_n = {\bigg[3 + 4n - 4\bigg]}^{2}$

$\rm :\longmapsto\:T_n = {\bigg[4n - 1\bigg]}^{2}$

$\rm :\longmapsto\:T_n = {16n}^{2} + 1 - 8n$

Now, We know that

$\rm :\longmapsto\:S_n = \displaystyle\sum_{n=1}^n \: T_n$

$\rm \:  =  \: \displaystyle\sum_{n=1}^n \: \bigg[ {16n}^{2} - 8n + 1 \bigg]$

$\rm \:  =  \: 16\displaystyle\sum_{n=1}^n {n}^{2} - 8\displaystyle\sum_{n=1}^n n + \displaystyle\sum_{n=1}^n 1$

$\rm \:  =  \: \dfrac{16n(n + 1)(2n + 1)}{6} - \dfrac{8n(n + 1)}{2} + n$

$\rm \:  =  \: \dfrac{8n(n + 1)(2n + 1)}{3} - 4n(n + 1) + n$

$\rm \:  =  \: n\bigg[\dfrac{8(n + 1)(2n + 1)}{3} - 4n - 4 + 1\bigg]$

$\rm \:  =  \: n\bigg[\dfrac{8(n + 1)(2n + 1)}{3} - 4n - 3\bigg]$

$\rm \:  =  \: n\bigg[\dfrac{8( {2n}^{2} + n + 2n + 1) - 12n - 9}{3}\bigg]$

$\rm \:  =  \: n\bigg[\dfrac{8( {2n}^{2} + 3n + 1) - 12n - 9}{3}\bigg]$

$\rm \:  =  \: n\bigg[\dfrac{{16n}^{2} + 24n + 8- 12n - 9}{3}\bigg]$

$\rm \:  =  \: n\bigg[\dfrac{{16n}^{2} + 12n - 1}{3}\bigg]$

Hence,

$\boxed{\tt{{3}^{2} + {7}^{2} + {11}^{2} + - - - n \: terms = \frac{n( {16n}^{2} + 12n - 1 )}{3} \: }}$

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$\boxed{\tt{ \displaystyle\sum_{n=1}^n 1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{n=1}^n n = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{n=1}^n {n}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\boxed{\tt{ \displaystyle\sum_{n=1}^n {n}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2} \: }}$

Answer 2

Given :

series

3² + 7² + 11² …+ n terms

To find :

Sum of the series

Solution :

n th terms of the given series

Tⁿ = [3 + (n - 1)4]²

= 16n² - 8n + 1

Sⁿ = ∑(16n² - 8n + 1)

= 16∑n² - 8∑n + ∑1

$\sf= 16 \lgroup \frac{n(n + 1)(2n + 1)}{6} \rgroup - 8 \lgroup \frac{n(n + 1)}{2} \rgroup + n$

$\sf = \frac{8}{3} n(n + 1)(2n + 1) - 4[n(n + 1)] + n$

= (4/3)n(n+1)(4n-1) + n

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The Required answer is

= (4/3)n(n+1)(4n-1) + n