Answer:
Option A,D
Explanation:
Concept involved
It is to convert into differences and using sum of n terms of AP
i.e, Sn=n2[2a+(n−1)d]
Sn=∑4nk=1(−1)k(k+1)2k2
=−(1)2−22+32+42−52−62+72+82+....
=(32−12)+(42−22)+(72−52)+(82−62)+....
= 2{(4+6+12+...) +(6+14+22+...)}
n terms n terms
=2[n2{2×4+(n+1)8}+n2{2×6+(n−1)8}]
=2[n(4+4n-4)+n(6+4n-4)]
=2[4n2+4n2+2n] =4n(n+1)
Here, 1056=32 x33, 1088=32 x34, 1120=32 x35, 1332=36x37
1056 and 1332 are possible answers