1. ## induction problem

Can anybody help me with this.

Prove that by mathematical induction for n>=1
1/1.4 + 1/4.7 + ................+ 1/(3n-2).(3n+1) = n/(3n+1)

2. Originally Posted by baz
Can anybody help me with this.

Prove that by mathematical induction for n>=1
1/1.4 + 1/4.7 + ................+ 1/(3n-2).(3n+1) = n/(3n+1)

RHS can be rewritten as $\displaystyle \frac13\left(1-\frac1{3n+1}\right)$.
$\displaystyle \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)$

3. Originally Posted by kompik
RHS can be rewritten as $\displaystyle \frac13\left(1-\frac1{3n+1}\right)$.
$\displaystyle \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)$
I didnot det you.How come R.H.S be rewritten as you said?

4. Originally Posted by baz
I didnot det you.How come R.H.S be rewritten as you said?
Just try to use the common denominator:
$\displaystyle \frac13\left(1-\frac1{3n+1}\right)= \frac13\left(\frac{3n+1}{3n+1}-\frac1{3n+1}\right)= \frac13\cdot\frac{3n}{3n+1}=\frac{n}{3n+1}$
which is exactly the RHS.
But the modification of the RHS is perhaps not so important.

Just try to use:
$\displaystyle \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)$
or (the same equation with n+1 instead of n
$\displaystyle \frac1{(3n+1)(3n+4)}=\frac13\left(\frac1{3n+1}-\frac1{3n+4}\right)$
in the inductive step of your proof.

EDIT: BTW the sums of this type are called telescopic. With a little practice you can recognize easily at least the simple ones - and then you see the result immediately (that was the reason why I rewrote the RHS - since I saw that this expression is what I get from the telescopic sum.

5. Originally Posted by kompik
Just try to use the common denominator:
$\displaystyle \frac13\left(1-\frac1{3n+1}\right)= \frac13\left(\frac{3n+1}{3n+1}-\frac1{3n+1}\right)= \frac13\cdot\frac{3n}{3n+1}=\frac{n}{3n+1}$
which is exactly the RHS.
But the modification of the RHS is perhaps not so important.

Just try to use:
$\displaystyle \frac1{(3n-2)(3n+1)}=\frac13\left(\frac1{3n-2}-\frac1{3n+1}\right)$
or (the same equation with n+1 instead of n
$\displaystyle \frac1{(3n+1)(3n+4)}=\frac13\left(\frac1{3n+1}-\frac1{3n+4}\right)$
in the inductive step of your proof.

EDIT: BTW the sums of this type are called telescopic. With a little practice you can recognize easily at least the simple ones - and then you see the result immediately (that was the reason why I rewrote the RHS - since I saw that this expression is what I get from the telescopic sum.