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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 options are
A Sn<π33
D Tn>π33
Sn<limnSn=limnnk=11n11+k/n+(k/n)2
=10dx1+x+x2=π33
Now, Tn>π33 as hn1k=0f(kh)>10f(x)dx>hnk=1f(kh)
Ans: A,D

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