Let Sn=n∑k=1nn2+kn+k2 and Tn=n−1∑k=0nn2+kn+k2 for n=1,2,3,…. Then,
A
Sn<π3√3
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B
Sn>π3√3
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C
Tn<π3√3
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D
Tn>π3√3
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Solution
The correct options are ASn<π3√3 DTn>π3√3 Sn<limn→∞Sn=limn→∞n∑k=11n⋅11+k/n+(k/n)2 =∫10dx1+x+x2=π3√3 Now, Tn>π3√3 as hn−1∑k=0f(kh)>∫10f(x)dx>hn∑k=1f(kh) Ans: A,D