# Help with arithmetic series.

• December 27th 2010, 02:17 PM
nelly1
Help with arithmetic series.
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.]

j
• December 27th 2010, 02:27 PM
Plato
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 $\sum\limits_{r = 0}^6 {r\left( {r + 1} \right)}$.
• December 27th 2010, 02:29 PM
snowtea
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,
$\sum_{r=0}^6 r(r+1) = 2\binom{8}{3} = 2\frac{8\times 7 \times 6}{3!} = 112$
• December 27th 2010, 02:35 PM
nelly1
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.