Among
(S1): lim n → ∞ 1/n²(2+4+6+… …+2 n)=1
(S2) : lim n → ∞ 1/n¹⁶(1¹⁵+2¹⁵+3¹⁵+… …+n¹⁵)=1/16

Solution

Maths sir · 3 steps

Pull out powers of n until each term looks like f(r/n), then the sum becomes an integral from 0 to 1.

1. Check S1. The bracket is 2(1+2+…+n) = 2 × n(n+1)/2 = n(n+1). Dividing by n² gives 1 + 1/n, which tends to 1. So S1 is true.
2. Rewrite S2. Split 1/n¹⁶ as 1/n × 1/n¹⁵. Each term then becomes 1/n(r/n)¹⁵, a Riemann sum for x¹⁵ on [0, 1].

   lim(n → ∞) 1/nΣᵣ₌₁ⁿ(r/n)¹⁵ = ∫₀¹ x¹⁵ dx = 1/16

3. So. Both limits match their stated values, so both statements are true, which is option (b).

The idea to keep. For 1/nᵏ⁺¹Σ rᵏ, write it as 1/nΣ (r/n)ᵏ. The limit is ∫₀¹ xᵏ dx = 1/(k+1).

Answer from the NTA final answer key. Solution by PrepWiser.