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Question: $\lim_{n \to n}\frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+...+\le...

lim⁡n→n3n{4+(2+1n)2+(2+2n)2+...+(3−1n)2}\lim_{n \to n}\frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+...+\left(3-\frac{1}{n}\right)^2\right\} is equal to:

A

12

B

193\frac{19}{3}

C

0

D

19

Answer

19

Explanation

Solution

The given limit is lim⁡n→∞3n{4+(2+1n)2+(2+2n)2+...+(3−1n)2}\lim_{n \to \infty}\frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+...+\left(3-\frac{1}{n}\right)^2\right\}.

We can rewrite the terms in the sum. The first term is 4=22=(2+0n)24 = 2^2 = \left(2+\frac{0}{n}\right)^2.

The terms are (2+0n)2,(2+1n)2,(2+2n)2,...,(2+n−1n)2\left(2+\frac{0}{n}\right)^2, \left(2+\frac{1}{n}\right)^2, \left(2+\frac{2}{n}\right)^2, ..., \left(2+\frac{n-1}{n}\right)^2.

The sum can be written as ∑i=0n−1(2+in)2\sum_{i=0}^{n-1} \left(2+\frac{i}{n}\right)^2.

The given limit becomes lim⁡n→∞3n∑i=0n−1(2+in)2\lim_{n \to \infty}\frac{3}{n} \sum_{i=0}^{n-1} \left(2+\frac{i}{n}\right)^2.

This limit is in the form of a Riemann sum. The general form of a definite integral as a limit of a sum is ∫abf(x)dx=lim⁡n→∞b−an∑i=0n−1f(a+ib−an)\int_a^b f(x) dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{i=0}^{n-1} f\left(a + i\frac{b-a}{n}\right) or ∫abf(x)dx=lim⁡n→∞b−an∑i=1nf(a+ib−an)\int_a^b f(x) dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\left(a + i\frac{b-a}{n}\right).

Let's match the given limit with the first form.

We have 3n∑i=0n−1(2+in)2\frac{3}{n} \sum_{i=0}^{n-1} \left(2+\frac{i}{n}\right)^2.

Compare this with b−an∑i=0n−1f(a+ib−an)\frac{b-a}{n} \sum_{i=0}^{n-1} f\left(a + i\frac{b-a}{n}\right).

We can see that f(a+ib−an)=(2+in)2f\left(a + i\frac{b-a}{n}\right) = \left(2+\frac{i}{n}\right)^2.

Let f(x)=x2f(x) = x^2.

Then f(a+ib−an)=(a+ib−an)2f\left(a + i\frac{b-a}{n}\right) = \left(a + i\frac{b-a}{n}\right)^2.

Comparing (a+ib−an)2\left(a + i\frac{b-a}{n}\right)^2 with (2+in)2\left(2+\frac{i}{n}\right)^2, we can set a=2a=2 and b−an=1n\frac{b-a}{n} = \frac{1}{n}.

From b−an=1n\frac{b-a}{n} = \frac{1}{n}, we get b−a=1b-a=1.

Since a=2a=2, we have b−2=1b-2=1, which gives b=3b=3.

The function is f(x)=x2f(x) = x^2 and the interval is [a,b]=[2,3][a, b] = [2, 3].

The term 3n\frac{3}{n} in the limit is 3×1n3 \times \frac{1}{n}.

The factor b−an\frac{b-a}{n} in the Riemann sum formula is 1n\frac{1}{n} in our case.

So the limit is equal to 3×∫abf(x)dx3 \times \int_a^b f(x) dx.

The integral is ∫23x2dx\int_2^3 x^2 dx.

∫x2dx=x33\int x^2 dx = \frac{x^3}{3}.

∫23x2dx=[x33]23=333−233=273−83=193\int_2^3 x^2 dx = \left[\frac{x^3}{3}\right]_2^3 = \frac{3^3}{3} - \frac{2^3}{3} = \frac{27}{3} - \frac{8}{3} = \frac{19}{3}.

The limit is 3×∫23x2dx=3×193=193 \times \int_2^3 x^2 dx = 3 \times \frac{19}{3} = 19.