Series

**Simplicity** · September 16th 2007, 06:06 AM

I need help with these four questions (If possible, can it be explained in steps):

1.
Work out $\displaystyle\sum_{r=n}^{2n}\frac{r^2}{1}$

2.
Given that $f(r) \equiv 1 \frac{1}{r(r+1)}$ , show that $f(r) - f(r+1) \equiv \frac{2}{r(r+1)(r+2)}$

3.
Prove that $\displaystyle\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)}= \frac{n}{2(n+2)}$

4.
Find the sum of the series $1^2 - 2^2 + 3^2 - 4^2 + ... - (2n)^2$

**CaptainBlack** · September 16th 2007, 06:27 AM

Originally Posted by Air

I need help with these four questions (If possible, can it be explained in steps):

1.
Work out $\displaystyle\sum_{r=n}^{2n}\frac{r^2}{1}$

I suppose that you know that:

$ \sum_1^n r^2 = \frac{n(n+1)(2n+1)}{6} $

So:

$ \sum_n^{2n} r^2 = $ $\sum_{r=1}^{2n}r^2 - \sum_{r=1}^{n-1} r^2$ $ = \frac{2n(2n+1)(4n+1)}{6}-\frac{(n-1)(n)(2(n-1)+1)}{6} $

and I leave the simplification of this to you.

RonL

**Krizalid** · September 16th 2007, 06:31 AM

Originally Posted by Air

3.
Prove that $\displaystyle\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)}= \frac{n}{2(n+2)}$

$\sum\limits_{r = 1}^n {\frac{1} {{\left( {r + 1} \right)\left( {r + 2} \right)}}} = \sum\limits_{r = 1}^n {\frac{1} {{r + 1}} - \frac{1} {{r + 2}}} = \frac{1} {{1 + 1}} - \frac{1} {{n + 2}} = \frac{n} {{2(n + 2)}}$

**CaptainBlack** · September 16th 2007, 06:38 AM

Originally Posted by Air

2.
Given that $f(r) \equiv 1 \frac{1}{r(r+1)}$ , show that $f(r) - f(r+1) \equiv \frac{2}{r(r+1)(r+2)}$

$ f(r) - f(r+1) = \frac{1}{r(r+1)}- \frac{1}{(r+1)(r+2)}=\frac{(r+2) - r}{r(r+1)(r+2)}=\frac{2}{r(r+1)(r+2)} $

RonL

**CaptainBlack** · September 16th 2007, 06:45 AM

Originally Posted by Air

4.
Find the sum of the series $1^2 - 2^2 + 3^2 - 4^2 + ... - (2n)^2$

Group pairs of terms:

$ \sum_{r=1}^{2n} r^2= \sum_{k=1}^{n} (2k-1)^2-(2k)^2=\sum_{k=1}^n (-4k+1)=n - 4 \sum_{k=1}^n k $

Which is a sum you should be able to do.

RonL

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