# Thread: Proof Question: Using Mathematical Induction

1. ## Proof Question: Using Mathematical Induction

i have been on this problem for a while and i dont get any where.

The problem statement, all variables and given/known data
Prove that, for all integers n=>1

i get to the point where i need to prove that S(k+1) also true but what is k+1?

is it 1/k(k+1)(k+3)

or (k+1)(k+3)/(k+1)(k+3)(k+5)

we had one lesson on this and this home work is damn hard :'(

thank you ever so much

2. Originally Posted by boomshine57th

i have been on this problem for a while and i dont get any where.

The problem statement, all variables and given/known data
Prove that, for all integers n=>1

i get to the point where i need to prove that S(k+1) also true but what is k+1?

is it 1/k(k+1)(k+3)

or (k+1)(k+3)/(k+1)(k+3)(k+5)

we had one lesson on this and this home work is damn hard :'(

thank you ever so much
Firstly, I think your equation to proove is wrong. I believe it should start:
$\frac{1}{2*4}+...$
Assuming that's right, we first need to proove it is true for n=1. A quick calculation is all that's required there - we find both sides are equal to $\frac{1}{8}$
Now we say "Assuming the statement is true for n=k, will it be true for n=k+1?" What we are assuming is the following:
$\sum_{r=1}^n \frac{1}{(r+1)(r+3)} = \frac{n(5n+13)}{12(n+2)(n+3)}$
What we are trying to show is that
$\sum_{r=1}^{n+1} \frac{1}{(r+1)(r+3)} = \frac{(n+1)[5(n+1)+13]}{12[(n+1)+2][(n+1)+3]}$
Now the RHS simplifies to $\frac{(n+1)(5n+18)}{12(n+3)(n+4)}$
To proove, rewrite the LHS: $\sum_{r=1}^n \frac{1}{(r+1)(r+3)} + \frac{1}{(n+2)(n+4)} = \frac{n(5n+13)}{12(n+2)(n+3)} + \frac{1}{(n+2)(n+4)}$

Now you just need to add the fractions, multiply out the numerator, re-factorise it and you should be there!

3. Thank You ever so much! you are a genius

thank you for going through all that trouble and writing it all out and thank you for noticing my mistake.

you are a life saver