Question

hey friend plz solve quest. 6 only

Answer 1

1).1³+2³+3³=+20³

➡1*1*1*+2*2*2+3*3*3=20*20*20

➡1+8+27+800

➡9+827

➡836

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

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