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Question

Let
Sn=nk=1nn2+kn+k2 and Tn=n1k=0nn2+kn+k2
for n=1,2,3,.. Then,

A
Sn<π33
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B
Sn>π33
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C
Tn<π33
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D
Tn>π33
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Solution

The correct option is C Tn<π33
Sn=nk=1nn2+kn+k2

Sn=1nnk=111+kn+(kn)2

Sn=1011+x+x2

Sn=π33

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