Method of Difference

**Punch** · June 10th 2011, 11:09 PM

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

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

**TheCoffeeMachine** · June 10th 2011, 11:28 PM

Sum ranges from r =1, not n =1, right? It's equivalent to:

$\sum_{r=-1}^{n-2}\frac{1}{r+2}+3\sum_{r=0}^{n-1}\frac{1}{r+2}-4\sum_{r=1}^{n}\frac{1}{r+2}$

**Punch** · June 11th 2011, 12:02 AM

Originally Posted by TheCoffeeMachine

Sum ranges from r =1, not n =1, right? It's equivalent to:

$\sum_{r=-1}^{n-2}\frac{1}{r+2}+3\sum_{r=0}^{n-1}\frac{1}{r+2}-4\sum_{r=1}^{n}\frac{1}{r+2}$

sorry, mistake on my part

**TheCoffeeMachine** · June 11th 2011, 12:46 AM

$\begin{aligned} S & = \sum_{1 \le r \le n}\frac{1}{r}+3\sum_{1 \le r \le n}\frac{1}{r+1}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \sum_{1 \le r+2 \le n}\frac{1}{r+2}+3\sum_{1 \le r +1\le n}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \sum_{-1 \le r \le n-2}\frac{1}{r+2}+3\sum_{0 \le r\le n-1}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \frac{1}{(-1)+2}+\frac{1}{(0)+2}-\frac{1}{(n-1)+2}-\frac{1}{(n)+2} +\frac{3}{(0)+2}-\frac{3}{(n)+2} \\& +\sum_{1 \le r \le n}\frac{1}{r+2}+3\sum_{1 \le r \le n}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = 3-\frac{1}{n+1}-\frac{4}{n+2}+4\sum_{1 \le r \le n}\frac{1}{r+2}-4\sum_{1 \le r \le n}\frac{1}{r+2} \\& = \boxed{3-\frac{1}{n+1}-\frac{4}{n+2}}.\end{aligned}$

**Soroban** · June 11th 2011, 05:11 AM

Hello, Punch!

$\text{Find: }\;S \;=\;\sum_{r = 1}^n \left(\frac{1}{r}+\frac{3}{r+1}-\frac{4}{r+2}\right)$

Write out the terms . . .

$S \;=\;\left(\frac{1}{1} + \frac{3}{2} - \frac{4}{3}\right) + \left(\frac{1}{2} + \frac{3}{3} - \frac{4}{4}\right) + \left(\frac{1}{3} + \frac{3}{4} - \frac{4}{5}\right) + \left(\frac{1}{4} + \frac{3}{5} - \frac{4}{6}\right) +\; \hdots$

. . . . $\hdots \;+ \left(\frac{1}{n-2} + \frac{3}{n-1} - \frac{4}{n}\right) + \left(\frac{1}{n-1} + \frac{3}{n} - \frac{4}{n+1}\right) + \left(\frac{1}{n} + \frac{3}{n+1} - \frac{4}{n+2}\right)$

We note that most of the terms cancel out . . .

$S \;=\;\left(\frac{1}{1} + \frac{3}{2} - \rlap{/}\frac{4}{3}\right) + \left(\frac{1}{2} + \rlap{/}\frac{3}{3} - \rlap{/}\frac{4}{4}\right) + \left(\rlap{/}\frac{1}{3} + \rlap{/}\frac{3}{4} - \rlap{/}\frac{4}{5}\right) + \left(\rlap{/}\frac{1}{4} + \rlap{/}\frac{3}{5} - \rlap{/}\frac{4}{6}\right) +\; \hdots$

. . . . $\hdots\; + \left(\rlap{/////}\frac{1}{n-2} + \rlap{/////}\frac{3}{n-1} - \rlap{/}\frac{4}{n}\right) + \left(\rlap{/////}\frac{1}{n-1} + \rlap{/}\frac{3}{n} - \frac{4}{n+1}\right) + \left(\rlap{/}\frac{1}{n} + \frac{3}{n+1} - \frac{4}{n+2}\right)$

And we have: . $S \;=\;\frac{1}{1} + \frac{3}{2} + \frac{1}{2} - \frac{4}{n+1} + \frac{3}{n+1} - \frac{4}{n+2}$

. . . . . . . . . . . $S \;=\;3 - \frac{1}{n+1} - \frac{4}{n+2} \;=\;\frac{3n^2+4n}{(n+1)(n+2)}$

Therefore:. $S \;=\;\frac{n(3n+4)}{(n+1)(n+2)}$

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