sum prove

**thereddevils** · February 18th 2009, 12:47 AM

If $U_n=\frac{1}{(2n+1)(2n+3)}$ , prove that

$<br /> U_{n-1}-U_n=\frac{4}{(2n-1)(2n+1)(2n+3)}<br />$

Thanks .

**Air** · February 18th 2009, 01:01 AM

Originally Posted by thereddevils

If $U_n=\frac{1}{(2n+1)(2n+3)}$ , prove that

$<br /> U_{n-1}-U_n=\frac{4}{(2n-1)(2n+1)(2n+3)}<br />$

Thanks .

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

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

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

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

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

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

Simplify the numerator.

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

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