Find the sum of the series 22 + 42 + 62 + 82 +................ 20 terms
Given series: 22 + 42 + 62 + 82 +.............
Nth term of the series
tn=(2n)2=4n2
Sum of the n terms
Sn=∑ni=1tn=4∑ni=1n2
=4n(n+1)(2n+1)6
On substituting n=20, we get-
=4×20×21×416=11480