1)

Let   Sn=4n2(1)k(k+1)2k2.Then , Sn can take value (s)


A) 1056

B) 1088

C) 1120

D) 1332

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+(n1)d]

    Sn=4nk=1(1)k(k+1)2k2

 =(1)222+32+425262+72+82+....

=(3212)+(4222)+(7252)+(8262)+....

= 2{(4+6+12+...)  +(6+14+22+...)}

    n terms                    n terms

=2[n2{2×4+(n+1)8}+n2{2×6+(n1)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