Series and Summations

**kleyzam** · October 26th 2008, 12:33 PM

I'm having a difficulty in understanding the below questions:

(1) Use the natural number series to find the following

(a) The sum of the first n terms in the series (1)(3)(5) + (2)(4)(6) + (3)(5)(7) + ...

(b) $\sum_{r=1}^{2n} r^2$

(c) $(n+1)^2 + (n+2)^2 + ... ... + (2n)^2$

(2) Expand $\frac{6}{p(p+3)}$ into partial fractions

(a) Use the above result to find $\sum_{r=1}^{n} \frac{6}{r(r+3)}$

(b) Hence find the value of $\frac{1}{4} + \frac{1}{10} + \frac{1}{18} + \frac{1}{28} ...$

**Rafael Almeida** · October 26th 2008, 01:06 PM

kleyzam,

For now I'll approach problem (2). I'm assuming you know the basics of expanding into partial fractions. If not, I suggest that you take a look in this material.

There must exist two numbers A and B such that

$\frac{6}{p(p+3)} = \frac{A}{p} + \frac{B}{p+3}$

Now multiply both sides by $p(p+3)$ :

$ \frac{6p(p+3)}{p(p+3)} = \frac{Ap(p+3)}{p} + \frac{Bp(p+3)}{p+3} $

$ \frac{6\rlap{///////}p(p+3)}{\rlap{///////}{p(p+3)}} = \frac{A\rlap{/}p(p+3)}{\rlap{/}p} + \frac{Bp\rlap{//////}(p+3)}{\rlap{/////}p+3} $

$ 6 = A(p+3) + Bp = p(A+B) + 3A $

Now you equate coefficients, and easily see that A = 2 and B = -2. So,

$\frac{6}{p(p+3)} = \frac{2}{p} - \frac{2}{p+3}$

This is the very same process you use when evaluating intrincate polynomial integrals. If you've gone past Calculus 101 you might remember this, otherwise save this knowledge as it'll help you later.

So, for (a):

$\sum_{r=1}^{n} \frac{6}{r(r+3)} = 2\sum_{r=1}^{n} \frac{1}{p} - 2\sum_{r=1}^{n} \frac{1}{p+3}$

As for (b), are we supposed to calculate the series sum into infinity?

**kleyzam** · October 26th 2008, 01:35 PM

Hi Rafael Almeida,

Thanks a lot for your reply, with regards to question b, i am assuming that the series sum up to n as in question (a)

**Rafael Almeida** · October 26th 2008, 03:03 PM

kleyzam,

If it sums up to n, then it's solved. If in this particular case you're supposed to calculate with $n\to\infty$ , then a different approach may be needed. Actually, notice that the series $\sum^\infty_n \frac{1}{n}$ in particular, diverges.

**Krizalid** · October 26th 2008, 03:37 PM

It's worth to mention that,

$\frac{6}{p(p+3)}=2\cdot \frac{p+3-p}{p(p+3)}=\frac{2}{p}-\frac{2}{p+3}.$

**Rafael Almeida** · October 26th 2008, 03:47 PM

Originally Posted by Krizalid

It's worth to mention that,

$\frac{6}{p(p+3)}=2\cdot \frac{p+3-p}{p(p+3)}=\frac{2}{p}-\frac{2}{p+3}.$

Sure, I mentioned this in my earlier post. Sorry to ask, but, what's the point?

PS: how did you do the partial fraction separation so fast?

**Krizalid** · October 26th 2008, 03:49 PM

Originally Posted by Rafael Almeida

Sorry to ask, but, what's the point?

The point is: partial fractions method is a safe way to descompose a ratio, the thing is that sometimes it's quite slow, 'cause when one can see a faster method, it's worth to make it work, since you save some of time.

Originally Posted by Rafael Almeida

PS: how did you do the partial fraction separation so fast?

Just look at factors of the denominator, I just used them against numerator.

**Rafael Almeida** · October 26th 2008, 03:53 PM

Originally Posted by Krizalid

The point is: partial fractions method is a safe way to descompose a ratio, the thing is that sometimes it's quite slow, 'cause when one can see a faster method, it's worth to make it work, since you save some of time.

Just look at factors of the denominator, I just used them against numerator.

Oh, now I see the 'trick'. Nice one, by the way.

**Soroban** · October 26th 2008, 06:52 PM

Hello, kleyzam!

(1) Use the natural number series to find the following

I assume this means that we can use these formulas:

. . $\sum^n_{k=1}k \;=\;\frac{n(n+1)}{2} \qquad\quad \sum^n_{k=1}k^2 \;=\;\frac{n(n+1)(2n+1)}{6} \qquad\quad \sum^n_{k=1}k^3 \;=\;\frac{n^2(n+1)^2}{4}$

(a) The sum of the first $n$ terms of: . $1\cdot3\cdot5 + 2\cdot4\cdot6 + 3\cdot5\cdot7 + \hdots$

We have: . $\sum^n_{k=1}k(k+2)(k+4) \;=\;\sum (k^3 + 6k^2 + 8k) \;=\;\sum^n_{k=1}k^3 + 6\sum k^2 + 8\sum k$

. . . . $= \;\frac{n^2(n+1)^2}{4} + {\color{red}\rlap{/}}6\!\cdot\!\frac{n(n+1)(2n+1)}{{\color{red}\rlap{/}}6} + {\color{red}\rlap{/}}8^4\!\cdot\!\frac{n(n+1)}{{\color{red}\rlap{/}}2}$

Factor: . $\frac{n(n+1)}{4}\bigg[n(n+1) +4(2n+1) +16\bigg] \;=\;\frac{n(n+1)}{4}(n^2+9n+20)$

. . . . . $S_n \;= \;\frac{n(n+1)(n+4)(n+5)}{4}$

$(b)\;\sum_{r=1}^{2n} r^2$

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

$(c)\;(n+1)^2 + (n+2)^2 + \hdots + (2n)^2$

This is the sum of the first $2n$ squares minus the sum of the first $n$ squares.

$S \;=\;\bigg[1^2+2^2+3^2 + \hdots + (n+1)^2 + (n+2)^2 + \hdots + (2n)^2\bigg] - \bigg[1^2+2^2+3^2 + \hdots + n^2\bigg]$

. . $= \;\frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6} \;=\;\frac{n(2n+1)}{6}\bigg[2(4n+1) - (n+1)\bigg]$

. . $= \;\frac{n(2n+1)(7n+1)}{6}$

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