IIT JEE , AIEEE Forum - Mathematics , Algebra

beethoven

find the sum of this series

(1²/1) + (1²+2²/1+2)+...............................+(1²+2²+3²+.....+n²/1+2+......+n)

plzzzzzzzzzzzzz

tango_goat

ans=sum of all squares of 1st n natural numbers/sum of 1st n natural numbers
=2*n(n-1)(2*n-1)/6n(n-1)
=(2*n-1)/3

vidyun_dusa

it is

(n(n+1)(2n+1)/6)/(n(n+1)/2)=

(2n+1)/3=(2/3

n)+(n/3)

=(2n²+3n)/3

nishant_88

according to me vidyun_dusa is correct

nadeemoidu

(2n+1)/3= (2/3)

n+n/3
              = n(n+1)/3 +n/3
              = (n² +2n)/3
              = n(n+2)/3

goutam.chalasani

dusa is wrong put n=1 in his solution it doesnot satisfy
nadeemoidu is correct

chandra007

calculate nth term=2n+1/3
now take sigma 1 to n of this

u get
=(n^2+3n)/3

vaibhav_hyderabad1991

(n(n+1)(2n+1)/6)/(n(n+1)/2)=(2n+1)/3=(2/3n)+(n/3)

=(2n2+3n)/3

wnnaiit

answer should be n(n+2)/3