Let Sn=∑nk=1nn2+kn+k2andTn=∑n−1k=0nn2+kn+k2 for a = 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 We have ∑nk=1nn2+kn+k2andTn=∑n−1k=0nn2+kn+k2n=1,2,3.....Forn=1wegetS1=11+1+1=13=0.3andT1=11+0=1Alsoπ3√3=π√39=3.14×1.739=0.34×1.73=0.58∴S1<π3√3<T1,∴Sn<π3√3andTn>π3√3