Question

`\lim _{n\to \infty }\sum _{i=1}^n\frac{3}{n}[(\frac{i}{n})^2+1]`.........??

Answer 1

`4`

Explanation:

`= lim_{n->oo} (3/n^3) [ sum_{i=1}^{i=n} i^2 ] + (3/n) [ sum_{i=1}^{i=n} 1 ]`
`"(Faulhaber's formula)"`
`= lim_{n->oo} (3/n^3) [ (n(n+1)(2n+1))/6 ] + (3/n) [ n ]`
`= lim_{n->oo} (3/n^3) [ n^3/3 + n^2/2 + n/6 ] + (3/n) [ n ]`
`= lim_{n->oo} [1 + ((3/2))/n + ((1/2))/n^2 + 3]`
`= lim_{n->oo} [ 1 + 0 + 0 + 3 ]`
`= 4`

Answer 2

`4`.

Explanation:

Here is another way to solve the Problem :

Recall that, `int_0^1f(x)dx=lim_(n to oo)sum_(i=1)^n1/nf(i/n)...(star)`.

`:." The Reqd. Lim.="lim_(n to oo)sum_(i=1)^n3/n{(i/n)^2+1}`,

`=3[lim_(n to oo)sum_(i=1)^n1/n{(i/n)^2+1}]`,

`=3int_0^1{(x)^2+1}dx............[because, (star)]`,

`=3[x^3/3+x]_0^1`,

`=[x^3+3x]_0^1`,

`=1^3+3xx1-(0^3+3xx0)`,

`rArr " The Reqd. Lim.="4`.