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