# Mathematical induction

• Aug 29th 2010, 02:07 PM
cozza
Mathematical induction
Hi, the question I have is:

a) Use mathematical induction to prove that

$\sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$ for n=1,2,3...

b) Deduce that

$\sum\frac{1}{r(r+2)}<\frac{3}{4}$ for n=1,2,3...

For a) I have the following, so far:

Let p(n) be the variable proposition $\sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$

For n=1, the sum is
$\frac{1}{1(1+2)}=\frac{1}{3}$

The right hand side gives:
$\frac{3}{4}-\frac{2*1+3}{2(1+1)(1+2)}=\frac{1}{3}$

So p(1) is tue.

Assuming p(k) is true, I get:
$\sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac {1}{15}+...+\frac{1}{k(k+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}$

next I tried to deduce that p(k+1) is true, that is:
$\sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac {1}{15}+...+\frac{1}{k(k+2)}+\frac{1}{(k+1)(k+3)}= \frac{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}$

Is this right so far?

• Aug 29th 2010, 02:11 PM
yeKciM
Quote:

Originally Posted by cozza
Hi, the question I have is:

a) Use mathematical induction to prove that

$\sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$ for n=1,2,3...

b) Deduce that

$\sum\frac{1}{r(r+2)}<\frac{3}{4}$ for n=1,2,3...

For a) I have the following, so far:

Let p(n) be the variable proposition $\sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$

For n=1, the sum is
$\frac{1}{1(1+2)}=\frac{1}{3}$

The right hand side gives:
$\frac{3}{4}-\frac{2*1+3}{2(1+1)(1+2)}=\frac{1}{3}$

So p(1) is tue.

Assuming p(k) is true, I get:
$\sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac {1}{15}+...+\frac{1}{k(k+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}$

next I tried to deduce that p(k+1) is true, that is:
$\sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac {1}{15}+...+\frac{1}{k(k+2)}+\frac{1}{(k+1)(k+3)}= \frac{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}$

Is this right so far?

yes :D

just now add that $\displaystyle \frac {1}{(k+1)(k+3)}$ to the $\displaystyle \frac {3}{4} -\frac {2k+3}{2(k+1)(k+2) }$
• Aug 29th 2010, 02:11 PM
Yes, you are summing from r=1 to r=n.
fine so far.
• Aug 30th 2010, 03:36 AM
cozza
Quote:

Originally Posted by yeKciM
yes :D

just now add that $\displaystyle \frac {1}{(k+1)(k+3)}$ to the $\displaystyle \frac {3}{4} -\frac {2k+3}{2(k+1)(k+2) }$

I am struggling with the algebra involved in doing the actual sum:

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

Are you able to give me any pointers? (Happy)
• Aug 30th 2010, 06:54 AM
Quote:

Originally Posted by cozza
I am struggling with the algebra involved in doing the actual sum:

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

Are you able to give me any pointers? (Happy)

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

$\displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r+2)}=\fra c{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}$

If you can show that P(k+1) will be definately true if P(k) is true,
then you will have completed the inductive step.

We use $\displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r+2)}=\sum _{r=1}^k\frac{1}{r(r+2)}+\frac{1}{(k+1)(k+3)}$

If P(k) is true, then we need to check if

$\displaystyle\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}=\fra c{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}$

The fraction 3/4 is irrelevant to comparing the LHS and RHS of the immediately above,
hence we want to know if

$\displaystyle\frac{1}{(k+1)(k+3)}-\frac{2k+3}{2(k+1)(k+2)}=-\frac{2k+5}{2(k+2)(k+3)}$

or

$\displaystyle\frac{2k+3}{2(k+1)(k+2)}-\frac{2}{2(k+1)(k+3)}=\frac{2k+5}{2(k+2)(k+3)}$

One way to check if the LHS equals the RHS is to multiply the 1st term of the LHS by $\displaystyle\frac{k+3}{k+3}$

and multiply the 2nd term of the LHS by $\displaystyle\frac{k+2}{k+2}$

and then multiply both sides (RHS and LHS) by $(k+1).$
Then both sides will have a common denominator, so you only need compare numerators.

If you prefer, you could post the steps you've taken and we can check your work.