3...
Tn=n(n+1)(2n+1)/6
Sn=ΣTn
thts all
How to find the sum of the following series-
(1)1/1x2+1/2x3+1/3x4+...upto n terms
(2)52+62+72+...202
(3)12+(12+22)+(12+22+32)+...upto n terms.
Please give proper methods.
2nd one , formula
n into n+1 into 2n+1 divided by 6
1st put n 20 than 4
then subtract 1 fm 2 !
what if in (3) instead of n terms we have 10 terms........then what will be the answer???
3...
if 10 is given
solve for n and substitute n as 10
TOO LENGHTY
use shortcuts
3.
The nth term will be Σn2
Therefore summation of the series will be ΣΣn2
Σn(n+1)(2n-1)/2
1/2Σ(n2+n)(2n-1)
1/2Σ2n3 + 2n2 - n2 -n
1/2Σ2n3 + n2 -n
n2(n + 1)2/2 - n(n+1)/2
This the answer to the third summation of series
n(n+1)/2 [ n(n+1) -1]