# sum prove

• Feb 18th 2009, 12:47 AM
thereddevils
sum prove
If $\displaystyle U_n=\frac{1}{(2n+1)(2n+3)}$, prove that

$\displaystyle U_{n-1}-U_n=\frac{4}{(2n-1)(2n+1)(2n+3)}$

Thanks .
• Feb 18th 2009, 01:01 AM
Air
Quote:

Originally Posted by thereddevils
If $\displaystyle U_n=\frac{1}{(2n+1)(2n+3)}$, prove that

$\displaystyle U_{n-1}-U_n=\frac{4}{(2n-1)(2n+1)(2n+3)}$

Thanks .

$\displaystyle U_n = \frac{1}{(2n+1)(2n+3)}$

$\displaystyle u_{n-1}$ is the point when considering a value one less than $\displaystyle n$ thus the expression is found (for $\displaystyle n-1$) when we substitute $\displaystyle n=n-1$.

$\displaystyle U_{n-1} = \frac{1}{(2(n-1)+1)(2(n-1)+3)} = \frac{1}{(2n-2+1)(2n-2+3)} = \frac{1}{(2n-1)(2n+1)}$

$\displaystyle \therefore U_{n-1} - U_n = \frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)}$

Bring the fraction into one by multiplying each fraction by common value.

$\displaystyle \frac{(2n+3) - (2n-1)}{(2n-1)(2n+1)(2n+3)}$

Simplify the numerator.

$\displaystyle \frac{4}{(2n-1)(2n+1)(2n+3)}$