Method of differences exam question [ ]

[thomas49th]

Method of differences exam question [urgent]

Homework Statement

$$\sum^{n}_{r=1}{\frac{5r+4}{r(r+1)(r+2)}$$

Now i know that this is equal to $$\frac{2}{r} + \frac{1}{r+1} -\frac{3}{r+2}$$

I need to use the method of differences to work out the summation of the series from 1 to n.

so i substitute values in

r=1:

$$2 + \frac{1}{2} - \frac{3}{3}$$

r=2:

$$1 + \frac{1}{3} - \frac{3}{4}$$

r=3:

$$\frac{2}{3} + \frac{1}{4} - \frac{3}{5}$$

r=n-1:

$$\frac{2}{n-1} + \frac{1}{n} - \frac{3}{n+1}$$

r=n:

$$\frac{2}{n} + \frac{1}{n+1} - \frac{3}{n+2}$$But I can't see any terms that have canceled out. How do i simplify this

Thanks :)
Thomas

 

[Dick]

r=n+1, (-3/(n+3))+1/(n+2)+2/(n+1). Look at the 2/(n+1) from this, the 1/(n+1) from r=n and the -3/(n+1) from r=n-1.

 

[Mark44]

To clarify, by "this" in your second line, you meant the summand, not the series sum.

You have $$\sum_{r = 1}^n (\frac{2}{r} + \frac{1}{r + 1} - \frac{3}{r + 2})$$
Break this sum into three separate summations. After that, move constant multiples outside the summation. Now, align the terms of all three summations by pulling out enough terms so that all of the sums start with the same fraction. I think you'll see that you get a lot of cancellation then.

 

[thomas49th]

Yup got it. Thanks a lot! diagonal 3!

Cheers :)