Quick Proof

**Misa-Campo** · August 1st 2008, 11:40 PM

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

**red_dog** · August 1st 2008, 11:46 PM

$1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}=1-\frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}=$

$=1+\frac{k+1-k-2}{(k+2)!}=1-\frac{1}{(k+2)!}$

**earboth** · August 1st 2008, 11:54 PM

Originally Posted by Misa-Campo

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

I assume that you know

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

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

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

which will yield the RHS.

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