Question

\(\lim_{n\to\infty} \frac{3}{n}\left\{4 + \left(2 + \frac{1}{n}\right)^2 + \left(2 + \frac{2}{n}\right)^2 + \dots + \left(3 - \frac{1}{n}\right)^2\right\}\) is equal to :

• A. 0
• B. \(\frac{19}{3}\)
• C. 19
• D. 12

Hint:

Write the terms as \(\left(2 + \frac{k}{n}\right)^2\) for \(k = 0, 1, \dots, n-1\) (the first term 4 is the \(k=0\) term) and convert the sum into a definite integral.

Answer 1

The Correct Option is C

The terms are \(\left(2 + \frac{k}{n}\right)^2\) for \(k = 0, 1, 2, \dots, n-1\): \(k=0\) gives 4 and \(k=n-1\) gives \(\left(3 - \frac{1}{n}\right)^2\).
So the limit is \(3\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}\left(2 + \frac{k}{n}\right)^2 = 3\int_0^1 (2 + x)^2\,dx\).
\(\int_0^1 (2+x)^2 dx = \left[\frac{(2+x)^3}{3}\right]_0^1 = \frac{27 - 8}{3} = \frac{19}{3}\).
Hence the limit \(= 3\cdot\frac{19}{3} = 19\).
Answer: (3) 19