How do i prove that the LHS=RHS? i am really confused because of factorials together with pronumerals

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- Aug 1st 2008, 11:40 PMMisa-CampoQuick 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 PMred_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 PMearboth
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.