
Induction Help
∑ i=1 to n, √[1+(1/i^2)+(1/(1+i^2))] = n(n+2)/n+1
First I did the base case of p(1) showing 3/2 on the LHS equals the 3/2 on the RHS.
Then I assumed p(k) and wrote out the formula with k in it.
Then prove p(k+1)= p(k)+ √[1+1/(k+1)^2+1/(k+2)^2]
=k(k+2)/k+1 + √[1+1/(k+1)^2+1/(k+2)^2]
Then I squared each to get rid of the square root.
(k(k+2)/(k+1))^2+ (k+1)^2/(k+1)^2 + 1/(k+1)^2 + 1/(k+2)^2
Now I'm stuck any Guidance would be great thanks!

Re: Induction Help
There are a couple of problems with your post. Firstly I don't believe that formula is correct. According to your formula the LHS when n=1 is $\displaystyle \sqrt{1+1+\frac{1}{2}}=\frac{\sqrt{10}}{2} \neq \frac{3}{2}.$
I believe what you should be trying to prove is
$\displaystyle \sum_{i=1}^{n}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i) ^2}}=\frac{n(n+2)}{n+1}.$
Secondly your approach for taking the inductive step seems a little strange in such proofs you prove the base case (i.e. n=1) and then assume the formula to be true for n=1,..,k. Then you consider the LHS when n=k+1, and use the fact that
$\displaystyle \sum_{i=1}^{k+1}f(i)=\sum_{i=1}^kf(i)+f(k+1)$
to prove the formula for all n.
I hope this helps.

Re: Induction Help
I'm just having trouble showing that, $\displaystyle \frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{k+1^2}+\frac{1}{(1+k+1)^2}}=\frac {k+1(k+1+2)}{k+1+1} $

Re: Induction Help
You've got to mess around a bit with the algebra:
$\displaystyle \sqrt{1+\frac{1}{i^2}+\frac{1}{(i+1)^2}}=\frac{1}{ i(i+1)}\sqrt{i^2(i+1)^2+(i+1)^2+i^2},$
Multiplying out the term in the square root and collecting like terms gives
$\displaystyle \frac{1}{i(i+1)}\sqrt{i^4+2i^3+3i^2+2i+1}=\frac{1} {i(i+1)}\sqrt{(i^2+i+1)^2}=\frac{i^2+i+1}{i(i+1)}$
The right hand side can then be rewritten as
$\displaystyle g(i)=\frac{i^2(i+2)(i1)(i+1)^2}{i(i+1)}.$
Now using $\displaystyle g(i)$ instead of $\displaystyle \sqrt{1+\frac{1}{i^2}+\frac{1}{(i+1)^2}}$ for $\displaystyle i=k+1$ you should be able to complete the induction.

Re: Induction Help
You might want to take a look at this topic:
http://mathhelpforum.com/algebra/209...tml#post758314
for a way to simplify your induction hypothesis.
edit: If I had taken the time to actually read the post above this one before posting, I would have seen the same simplification has already been given.(Giggle)