Show that there is exactly one positive integer n for which $\sum_{r=1}^n r^3+\sum_{r=1}^n r = 8 \sum_{r=1}^n r^2$

Question

Can someone show me the working for this? Thanks

AQA A Level Mathematics Further Pure 1 January 2010 Question 8(b) http://filestore.aqa.org.uk/subjects/AQA-MFP1-W-QP-JAN10.PDF

And if anyone could explain to me how to type equations properly instead of using text I would really appreciate that, Regards

The page for LaTeX Markup on StackExchange can be found here: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — Axoren, Feb 04 '15 at 01:13
Hint: Write down the standard closed forms for $\sum_1^n r$ and the other two. Now rewrite your equation using these. — André Nicolas, Feb 04 '15 at 01:25

score 2 · Accepted Answer · answered Feb 04 '15 at 01:28

2

In general,

$$\sum_{r=1}^n r = \frac{n(n+1)}{2},$$

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

and

$$\sum_{r=1}^n r^3 = \frac{n^2(n+1)^2}{4}.$$

Adding the first two gives us

$$\sum_{r=1}^n r^3 + \sum_{r=1}^n r = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)}{2} = \frac{n(n+1)(n^2+n+2)}{4}.$$

So then if

$$\sum_{r=1}^n r^3 + \sum_{r=1}^n r = 8 \sum_{r=1}^n r^2,$$

we have

$$\frac{n(n+1)(n^2+n+2)}{4} = 8\frac{n(n+1)(2n+1)}{6}.$$

Can you see how to proceed?

answered Feb 04 '15 at 01:28

Cameron Williams

I don't see how to proceed – Thomas Winkworth Feb 04 '15 at 01:52
You have an equation with a variable on both sides. Think of it in terms of algebra. You want to solve for $n$. – Cameron Williams Feb 04 '15 at 01:57
I got ya, thanks – Thomas Winkworth Feb 04 '15 at 02:03

1 Answers1