# Quick Proof

• Aug 1st 2008, 11:40 PM
Misa-Campo
Quick Proof
How do i prove that the LHS=RHS? i am really confused because of factorials together with pronumerals

http://img361.imageshack.us/img361/8...ductioncb5.gif
• Aug 1st 2008, 11:46 PM
red_dog
$\displaystyle 1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}=1-\frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}=$

$\displaystyle =1+\frac{k+1-k-2}{(k+2)!}=1-\frac{1}{(k+2)!}$
• Aug 1st 2008, 11:54 PM
earboth
Quote:

Originally Posted by Misa-Campo
How do i prove that the LHS=RHS? i am really confused because of factorials together with pronumerals

http://img361.imageshack.us/img361/8...ductioncb5.gif

I assume that you know

$\displaystyle (k+2)! = (k+1)! \cdot (k+2)$

The common denominator of the fractions therefore is (k+2)!

$\displaystyle 1-\frac1{(k+1)!} + \frac{k+1}{(k+2)!} = 1-\frac{k+2}{(k+1)! \cdot (k+2)} + \frac{k+1}{(k+2)!}$ $\displaystyle \ =\$ $\displaystyle 1-\left(\frac{k+2}{(k+1)! \cdot (k+2)} - \frac{k+1}{(k+2)!}\right) = 1-\frac{(k+2)-(k+1)}{(k+2)!}$

which will yield the RHS.