Question

find the sum to n terms of the series 1^2 + (1^2 + 2^2) + (1^2 + 2^2 +3^2) +...​

Answer 1

Given that,

$\sf a_n = \dfrac{n( {n}^{2} + 5) }{4}$

n = 1,2,3,4,5...

1st term

$\sf a_1= \dfrac{1( {1}^{2} + 5) }{4}$

$\sf a_1= \dfrac{6 }{4}$

$\sf a_1= \dfrac{3 }{2}$

similarly

second term,

$\sf a_2= \dfrac{2( {2}^{2} + 5) }{4}$

$\sf a_2= \dfrac{2(9) }{4}$

$\sf a_2= \dfrac{9}{2}$

Third term,

$\sf a_3= \dfrac{3( {3}^{2} + 5) }{4}$

$\sf a_3= \dfrac{3( 14) }{4}$

$\sf a_3= \dfrac{21}{4}$

Fourth term,

$\sf a_4= \dfrac{4( {4}^{2} + 5) }{4}$

$\sf a_4= \dfrac{4(21)}{4}$

$\sf a_4= 21$

Fifth term,

$\sf a_5= \dfrac{5( {5}^{2} + 5) }{4}$

$\sf a_5= \dfrac{5(30) }{4}$

$\sf a_5= \dfrac{75}{2}$

Therefore 1st five terms are 3/2, 9/2, 21/2, 21 and 75/2

Answer 2

Step-by-step explanation:

Tk=kth Term = k(k+1)(2k+1)/6.

Sn= Sum of first n terms of this sequence

= [Sum from k=1 to k=n] (2k^3+ 3k^2+k)/6

= ((n(n+1))^2/12) + ((n(n+1)(2n+1)/12) + (n(n+1)/12)

= (n(n+1)/12)× ((n^2)+n+2n+1+1)

= (n(n+1)/12)× (n+1)×(n+2)

= ( (n×((n+1)^2)×(n+2))/12 ) Answer.