Help with arithmetic series.

**nelly1** · December 27th 2010, 02:17 PM

Hi, I'm taking AS Maths Core 1 in January. I am stuck on a question in the text book.
It goes like this:
Calculate:

5
Sigma r(r+1)
r = 0

The answer in the text book is 112 but I can't get that. I get 70 as an answer. I would really appreciate any help with how to solve this particular problem.]

Thanks in advance,

j

**Plato** · December 27th 2010, 02:27 PM

The answer to the problem you posted is 70.

So check the problem again. Is it correct?

If so, the given answer is wrong. It goes with $\sum\limits_{r = 0}^6 {r\left( {r + 1} \right)}$ .

**snowtea** · December 27th 2010, 02:29 PM

The easiest way to do this summation is by using combinations:

$\sum_{r=0}^5 r(r+1) = \sum_{r=0}^5 2\binom{r+1}{2} = 2\binom{7}{3} = 2\frac{7\times 6 \times 5}{3!} = 70$

You can always check by a direct sum
0*1 + 1*2 + 2*3 + 3*4 + 4*5 + 5*6
= 0 + 2 + 6 + 12 + 20 + 30 = 70

so you are absolutely correct

However,
$\sum_{r=0}^6 r(r+1) = 2\binom{8}{3} = 2\frac{8\times 7 \times 6}{3!} = 112$

**nelly1** · December 27th 2010, 02:35 PM

Thanks so much. It had my head chewed up all day. It is a misprint in the book. I checked so many things except for the upper limit.

Thread: Help with arithmetic series.

LinkBack

Thread Tools

Display

Help with arithmetic series.

Similar Math Help Forum Discussions

Arithmetic Series Help

Arithmetic Progression or Arithmetic Series Problem

Arithmetic Series

Arithmetic series

arithmetic series

Search Tags