Question

find the sum of series whose nth term is (4n^3 + 6n^2 +2n)

Answer 1

if nth term is the above thus the first term would be
4(1)^3+6(1)^2+2(1)
=12
similarly we will get second term =
4(2)^3+6(2)^2+2(2)
=32+24+4
=60
then from subtracting A1 from a2
we will get d= 48
now as sn = n/2(2a+(n-1)d)
so we will just substitute the values and the answer in terms of n
n/2(24+(n-1)48)

but you will get a and d from here which will be used in sum gotcha??