1. ## Series

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 $1^2 - 2^2 + 3^2 - 4^2 + ... - (2n)^2$

2. 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

3. Originally Posted by Air
3.
Prove that $\displaystyle\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)}= \frac{n}{2(n+2)}$
$\sum\limits_{r = 1}^n {\frac{1}
{{\left( {r + 1} \right)\left( {r + 2} \right)}}} = \sum\limits_{r = 1}^n {\frac{1}
{{r + 1}} - \frac{1}
{{r + 2}}} = \frac{1}
{{1 + 1}} - \frac{1}
{{n + 2}} = \frac{n}
{{2(n + 2)}}$

4. 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

5. 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