lim(n → ∞) 3/n{4+(2+1/n)²+(2+2/n)²+…+(3-1/n)²} is equal to :

Solution

Maths sir · 4 steps

Write each term as (2+r/n)², let 1/n become dx, and the sum is an integral.

1. Spot the pattern. The first term is 4 = (2+0/n)² and the last is (3-1/n)² = (2+(n-1)/n)². So the bracket is Σᵣ₌₀ⁿ⁻¹(2+r/n)², which has n terms.
2. Sum to integral. Replace r/n by x and 1/n by dx. As r runs from 0 to n-1, x runs from 0 to 1. The factor 3 stays outside.

   lim(n→∞) 3/nΣᵣ₌₀ⁿ⁻¹(2+r/n)² = 3∫₀¹ (2+x)² dx

3. Evaluate. 3∫₀¹ (2+x)² dx = 3[(2+x)³/3]₀¹ = 3³ - 2³ = 27 - 8 = 19.
4. So. The limit equals 19, which is option (c).

The idea to keep. For a limit of a sum, write the general term as f(r/n), read the limits of x from the first and last values of r/n, and keep any extra constant outside.

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