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Question

If Sn=4n1(1)k(k+1)2k2. Then, Sn can take value(s)

A
1056
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B
1088
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C
1120
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D
1332
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Solution

The correct options are
A 1056
D 1332
Convert it into differences and use sum of n terms of an AP,
ie., Sn=n2[2a+(n1)d]
Now, Sn=4nk=1(1)k(k+1)2.k2
=(1)222+32+425262+72+82+
=(3212)+(4222)+(7252)+(8262)+
=2{(4+12+20)n terms+(6+14+22+)}n terms
=2[n2{2×4+(n1)8}+n2{2×6+(n1)8}]
=2[n(4+4n4)+n(6+4n4)]
=2[4n2+4n2+2n]=4n(4n+1)
Here, 1056=32×33,1088=32×34,
1120=32×35,1332=36×37
1056 and 1332 are possible answers

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