Question

If here is anyone who can solve my question correctly I challenge you 9^2 + 11^2 +13^2+...................to n terms if anyone can solve I think that person genius

Answer 1

We have a formula of sum of squares of odd nos=n(n+1)(2n+1)/6

But here is cannot substitute n...

Because the series starts from 9

Therefore for e for the above series ..

It would be the total sum of squares of all the n odd nos -(1^2+3^2+5^2+7^2)

=>

But the number of terms is also not simply n....but it would be n+4

By substituting

We get ...

((n+4)(n+5)(2n+5)/6)-(4(4+1)(2(4)+1)/6)

((n+4)(n+5)(2n+5)/6)-(4(5)(9)/6)

((n+4)(n+5)(2n+5)/6)-(30)

Hope it helps ....

Pls mark me as brainliest plsss

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1. Sum of the first n natural numbers :

$\displaystyle \: 1 + 2 + 3 + .... + n = \frac{n(n + 1)}{2}$

2. Sum of the squares of first n natural numbers :

$\displaystyle \:{1}^{2} + {2}^{2} + {3}^{2} + .... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6}$

TO DETERMINE

${9}^{2} + {11}^{2} + {13}^{2} + .........upto \: \: \: \: n \: terms$

EVALUATION

Let s be the sum and $t_n$be the n th term

Then

$t_n = { \{ \: n \: th \: \: \: term \: of \: the \: AP \: \: 9 \: , 11 \: , 13 , .......\}}^{2}$

$\implies \: t_n = { \{9 + (n - 1) \times 2 \}}^{2}$

$\implies \: t_n = { (2n + 7) }^{2}$

$\implies \: t_n = 4 {n}^{2} + 28n + 49$

Putting n = 1, 2, 3,....... n we get

$\: t_1= 4 \times {1}^{2} + 28 \times 1+ 49$

$\: t_2= 4 \times {2}^{2} + 28 \times 2+ 49$

$\: t_3= 4 \times {3}^{2} + 28 \times 3+ 49$

.

.

.

.

$\: t_n= 4 \times {n}^{2} + 28 \times n+ 49$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

On addition

$\: t_1 +t_2 +t_3 + ... + t_n = 4 \times ( {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} ) + 28 \times (1 + 2 + 3 + ... + n )+ 49n$

$\displaystyle \: \implies \: s \: = 4 \times \frac{n(n + 1)(2n + 1)}{6} + 28 \times \frac{n(n + 1)}{2} + 49n$

$\displaystyle \: \implies \: s \: = 2 \times \frac{n(n + 1)(2n + 1)}{3} + 14 \times {n(n + 1)} + 49n$

$\displaystyle \: \implies \: s \: = 2n(n + 1) \{ \frac{2n + 1}{3} + 7 \}+ 49n$

$\displaystyle \: \implies \: s \: = 2n(n + 1)( \frac{2n + 22}{3})+ 49n$

$\displaystyle \: \implies \: s \: = 4n(n + 1)( \frac{n + 11}{3})+ 49n$

$\displaystyle \: \implies \: s \: = n( \frac{ 4{n}^{2} + 48n + 44 + 147 }{3} )$

$\displaystyle \: \implies \: s \: = n( \frac{ 4{n}^{2} + 48n + 191 }{3} )$

$\displaystyle \: \implies \: s \: = \frac{ 4{n}^{3} + 48 {n}^{2} + 191n }{3}$

RESULT

The required sum is

$\displaystyle \: = \frac{ 4{n}^{3} + 48 {n}^{2} + 191n }{3}$