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so I'm struggling with this one question or better any question of this type:

You are given that $$\frac{3}{(5+3x)(2+3x)}\equiv \frac{1}{2+3x}-\frac{1}{5+3x}$$ use this result to find $$\sum_{r=1}^n \frac{1}{(5+3x)(2+3x)}$$ where n=100, giving your answer as an exact fraction.

I would know exactly what to do if they asked for the $$\sum_{r=1}^n \frac{3}{(5+3x)(2+3x)}$$ because then I would simply use the method of differences with $$\sum_{r=1}^n (\frac{1}{2+3x}-\frac{1}{5+3x})$$ but here I have no clue. I looked through all my textbooks and revisionguides etc. but the examples in there are only of the kind I know how to do. Please help I have only two weeks until my exam and I'm kinda freaking out...

Elvendoork
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2 Answers2

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You are given that $$ \frac{3}{(5+3x)(2+3x)}=\frac{1}{2+3x}-\frac{1}{5+3x}. $$ Therefore, $$ \sum_{x=1}^N\frac{1}{(5+3x)(2+3x)}=\frac{1}{3}\sum_{x=1}^N\frac{3}{(5+3x)(2+3x)} $$ At this point, we can substitute to get $$ \frac{1}{3}\sum_{x=1}^N\left(\frac{1}{2+3x}-\frac{1}{5+3x}\right). $$

Observe that $$ \sum_{x=1}^N\left(\frac{1}{2+3x}-\frac{1}{5+3x}\right)=\sum_{x=1}^N\frac{1}{2+3x}-\sum_{x=1}^N\frac{1}{5+3x} $$ Using the substitution $y=x+1$ on the second sum gives us $$ \sum_{x=1}^N\frac{1}{2+3x}-\sum_{y=2}^{N+1}\frac{1}{5+3(y-1)}=\sum_{x=1}^N\frac{1}{2+3x}-\sum_{y=2}^{N+1}\frac{1}{2+3y} $$ Observe that everything cancels except when $x=1$ and $y=N+1$, so you get $$ \frac{1}{5}-\frac{1}{5+3N}. $$

Michael Burr
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The series is

$\sum_{x=1}^N \frac{1}{(5+3x)(2+3x)} = \sum_{x=1}^N \frac{1}{2+3x}-\frac{1}{5+3x}$

$= \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + ... + \frac{1}{2+3N} - \frac{1}{5 +3N}$

All terms in the middle cancel out (called telescoping) as you can see.

Therefore, LHS $= \frac{1}{5} - \frac{1}{5 +3N}$

sku
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