The sum of the first n terms of the series 12+2.22+32+2.42+52+. Is n(n+1)22, when n is even, when 'n' is odd sum is
n(n+1)22
n2(n+1)2
n(n+1)24
3n(n+1)2
If n is odd, the required sum is
12+2.22+32+2⋅42+52+..+2(n−1)2+n2=(n−1)n22+n2=n2(n+1)2