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Question

Let Sn=nk=1nn2+kn+k2 and Tn=n1k=0nn2+kn+k2 for a = 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 options are
A Sn<π33
D Tn>π33
We have nk=1nn2+kn+k2and Tn=n1k=0nn2+kn+k2n=1,2,3.....For n=1 we getS1=11+1+1=13=0.3 and T1=11+0=1Also π33=π39=3.14×1.739=0.34×1.73=0.58S1<π33<T1,Sn<π33 and Tn>π33

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