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Ive been looking at this problem and trying to use examples online to try to solve it but I get stuck.

It says to use mathematical induction to prove (1/1*4)+(1/4*7)+(1/7*10)+ ... + (1/(3n-2)(3n+1)) = n/(3n+1)

I solve for n=1 and substitute k for n, but I don’t really know what to do after that step.

It would really help to see this problem laid out step by step. Thank you!!

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if $$ \sum_{j=1}^k \frac 1{(3j-2)(3j+1) } = \frac k {3k+1} $$ then $$ \sum_{j=1}^{k+1} \frac 1{(3j-2)(3j+1) } = \frac k {3k+1} + \frac 1{(3(k+1)-2)(3(k+1)+1) } $$

$$ = \frac k {3k+1} + \frac 1{(3k+1)(3k+4) } $$

$$ = \frac { k(3k+4) +1 }{(3k+1)(3k+4) } $$

$$ = \frac { 3k^2+4k+1 }{(3k+1)(3k+4) } $$

$$ = \frac { (3k+1)(k+1) }{(3k+1)(3k+4) } $$

$$ = \frac { k+1}{3k+4 } = \frac { (k+1)}{3(k+1) +1 } $$

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1) $n = 1$. Then $\frac 1{1*4}= \frac 1{3*1+1}$

2) assume true for $n$. Show true for $n+1$:

Then $\frac 1{1*4} + ... + \frac 1{(3n-2 )(3n+1)} = \frac n{3n+1}$ so

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

So all you have to do is show $\frac n{3n+1}+ \frac 1{3(n+1)-2)(3(n+1) +1)}= \frac {n+1}{3(n+1) + 1}$.

Can you do that?

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  • $\begingroup$ Yes, thanks for the help! $\endgroup$ – C Miller Dec 4 '17 at 0:50

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