1. ## [SOLVED] Proving equality - help please..

Hey guys. This is proberbly a bit simple, but im really tired and i have to solve this soon.

I have this equation:

(k+1)/(2(k+1)+1) = k/(2k+1) + 1/(4(k+1)^2 -1)

Its part of a proof by induction. I dont know if this is the correct forum, but im guessing its a pretty simple problem, im just too tired to solve it. Id just like someone to show me how to move the digits around to proove that this equality is true.

Thanks for the help.

Morten

2. Originally Posted by MortenDK
Hey guys. This is proberbly a bit simple, but im really tired and i have to solve this soon.

I have this equation:

(k+1)/(2(k+1)+1) = k/(2k+1) + 1/(4(k+1)^2 -1)

Its part of a proof by induction. I dont know if this is the correct forum, but im guessing its a pretty simple problem, im just too tired to solve it. Id just like someone to show me how to move the digits around to proove that this equality is true.

Thanks for the help.

Morten

Let's hope you remember your high school algebra, in particular squares difference: $a^2-b^2=(a-b)(a+b)$:

$\frac{k}{2k+1}+\frac{1}{4(k+1)^2-1}=\frac{k}{2k+1}+\frac{1}{\left[2(k+1)-1\right]\left[2(k+1)+1\right]}$ $=\frac{1}{2k+1}\left[k+\frac{1}{2k+3}\right]=\frac{1}{2k+1}\,\frac{2k^2+3k+1}{2k+3}$ $=\frac{1}{2k+1}\,\frac{(2k+1)(k+1)}{2k+3}$...and we're done.

Tonio

3. Thanks.. actually i solved it a couple of hours later. Took a break and then it all got a lot easier.
But thanks a bunch for the assistance