The correct option is C Sum of first ten terms is 5054
We have, S=1+1(1+3)(1+2)2+1(1+3+5)(1+2+3)2+1(1+3+5+7)(1+2+3+4)2+⋯
So, the general term can be written as
Tr=1r2(1+2+⋯+r)2=1r2(r(r+1)2)2
=(r+1)24=r2+2r+14
∴T7=16
And, T9=25
S10=10∑r=1r2+210∑r=1r+10∑r=114
=14[10×11×216+2(10×112)+10]
=5054