Question

Let Sn=∑k=01nn2+kn+k2 and Tn=∑k=01−1nn2+kn+k2, for n=1,2,3,… Then

• A. (A) sn<π33
• B. (B) Sn>π33
• C. (C) Tn<π33
• D. (D) Tn>π33

Answer 1

Answer: (A), (D)

(A) sn<π33

Answer 2

Explanation:
Given: sn=∑i=01nn2+kn+k2=∑k=011n(11+kn+k2n2)⟨lim1→∞∑k=011n(11+kn+(kn)2)=∫0111+x+x2dx=[23tan−1⁡(23(x+12))]01=23⋅(π3−π6)=π33=23⋅(π3−π6)=π33i.e. Sn<π33Similarly, Tn>π33