# Induction Help

1. Dec 10, 2012

### andyk23

1. The problem statement, all variables and given/known data

∑ i=1 to n1+(1/i2)+(1/(1+i)2)−−−−−−−−−−−−−−−−−−−−√ = n(n+2)/n+1

2. The attempt at a solution

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!

2. Dec 10, 2012

### SammyS

Staff Emeritus
The above line is nonsensical. Not enough parentheses, probably missing exponentiation, dangling radical ...

Do you mean $\displaystyle \sum_{i=1}^{n}\sqrt{1+1/(i^2)+1/(1+i)^2}\ \ = \ \ n(n+2)/(n+1)\ ?$

If so, then I must be a much better mind reader than I have ever been given credit for !

3. Dec 10, 2012

### andyk23

4. Dec 10, 2012

### SammyS

Staff Emeritus

5. Dec 10, 2012

### andyk23

Yes I meant that!

Last edited by a moderator: Dec 10, 2012
6. Dec 10, 2012

### SammyS

Staff Emeritus
As you stated: For the induction step, you assume that
$\displaystyle \sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ = \ \frac{k(k+2)}{k+1}$​
for k ≥ 1 . From that assumption, show that it follows that
$\displaystyle \sum_{i=1}^{k+1}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ = \ \frac{(k+1)(k+3)}{k+2}\ .$​

So look at $\displaystyle \ \ \sum_{i=1}^{k+1}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\,,\ \$ which you can write as $\displaystyle \ \ \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}+\sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ .$

Now use you assumption, then do some algebra.

7. Dec 11, 2012

### andyk23

I'm just having a trouble showing, $\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\ .$ = $\frac{k+1(k+1)}{k+1+2}\ .$

8. Dec 11, 2012

I'm just having a trouble showing, $\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}+\sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ .$ = $\frac{k+1(k+1+2)}{k+1}\ .[itex] 9. Dec 11, 2012 ### andyk23 I'm just having a trouble showing, [itex]\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\$ = $\frac{k+1(k+1+2)}{k+1}\$

10. Dec 11, 2012

### Michael Redei

What you really need to show next is
$$\frac{k(k+2)}{k+1}+\sqrt{1+\frac1{(k+1)^2}+\frac1{(k+2)^2}} = \frac{\color{red}{(k+1)} (k+3)}{k+\color{red}2}.$$
I'd try subtracting $\frac{k(k+2)}{k+1}$ on both sides and then squaring the results.