Proving sum of nested radicals by induction

[andyk23]

Homework Statement

$\sum$ i=1 to n$\sqrt{1+(1/i^2)+(1/(1+i)^2)}$ = 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)+ $\sqrt{1+1/(k+1)^2+1/(k+2)^2}$
=k(k+2)/k+1 + $\sqrt{1+1/(k+1)^2+1/(k+2)^2}$
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
Then I factored everything out
((k^4+4k^3+4k)/(k+1)^2) + ((k^2+2k+1)/(k+1)^2) + 1/(k+1)^2 + 1/(k+2)^2.
Basically I'm having a brain freeze on how to get (k+2) as the common denominator. Any Guidance would be great thanks!

 

[micromass]

=k(k+2)/k+1 + $\sqrt{1+1/(k+1)^2+1/(k+2)^2}$
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

So what you basically did was

$$(a+b)^2=a^2+b^2$$

This formula is not valid! And this is also what causes your mistake.

 

[andyk23]

Sorry I'm not following.. I understand what you're saying I did, just confused on which part I did the (a+b)^2 = a^2 + b^2

 

[micromass]

Correct me if I'm wrong, but you wanted to square

$$\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}$$

So it seems to me that you did the following:

$$\begin{eqnarray*}<br /> \left(\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\right)^2<br /> & = & \left(\frac{k(k+2)}{k+1}\right)^2 + \left(\sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\right)^2\\<br /> & = & \frac{k^2(k+2)^2}{(k+1)^2} + 1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}\\<br /> & = & \frac{k^2(k+2)^2}{(k+1)^2} + \frac{(k+1)^2}{(k+1)^2}+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}<br /> \end{eqnarray*}$$

Is this what you you were trying to do?? In that case, it is the first equality that is not correct.

 

[andyk23]

Sorry I had it written down on my paper but I didn't type it right! I'm just having a brain freeze on what I need to multiply the other rationals to have a denominator of (k+2)^2. Assuming that's the correct next step.