Question

Evaluate: lim_{n rightarrow infty}sum_{k=1}^{n}(-1+9k^{2}div n ^{2})*3div n
I just need help with the steps. (PLEEEZZEE)

Answer 1

\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(-1+9k^{2}\div n ^{2})*3\div n\]

Answer 2

and how do I evaluate that?

Answer 3

I think it's best to first see if it is convergent :-D

Answer 4

First find an expression for \( \sum_1^n k^2 \). Hence you can eliminate the sum and k from the expression altogether and just have an expression in n.

Answer 5

I think there is an expression for that, but I forgot what it was :-P let's google

Answer 6

I'll tell you right now that
\[ \sum_1^n k^2 = \frac{1}{6}n(n+1)(2n+1) \]

Using that, eliminate k from your expression.

Answer 7

\[\frac{n(n+1(2n+1)}{6}\]

Answer 8

aww beat me to it :) alright how do we fit that ugly fraction into that

Answer 9

I think we should break that sum apart

Answer 10

\[ \sum_{k=1}^n \frac{-1+9k^2 }{n^2} \frac{3}{n} = \frac{3}{n^3} \left( -\sum_{k=1}^n 1 + \sum_{k=1}^n 9k^2 \right) \]

Now you kwenisha, take it from here.

Answer 11

so no we have:
\[ 3divn(\sum_{k=1}^{n}-1+9divn \sum_{k=1}^{n}k ^{2}) => 3divn(-n+3divn(n+1)(2n+1)\div2\], this then give us: \[3divn(-n+3(n+1)(2n+1))\div2n\] and on and on and on... think I got the general idea :)

Answer 12

So *now* we have... not *no* :)

Answer 13

Be careful. You've already at least one mistake I can see.

Answer 14

oops...

Answer 15

Quick question: is this the general formula/equation when solving a problem like this n(n+1(2n+1)/6?