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- Homework Statement
- Show that ##\dfrac{1}{r(r+1)(r+2)} -\dfrac{1}{(r+1)(r+2)(r+3)}= \dfrac{3}{r(r+1)(r+2)(r+3)}##; hence sum the series ## \sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)(r+3)}##

- Relevant Equations
- Method of difference

Attempt;

##\dfrac{1}{r(r+1)(r+2)} -\dfrac{1}{(r+1)(r+2)(r+3)}=\dfrac{(r+3)-1(r)}{r(r+1)(r+2)(r+3)}=\dfrac{3}{r(r+1)(r+2)(r+3)}##

Let ##f(r)=\dfrac{1}{r(r+1)(r+2)}##

##f(r+1)= \dfrac{1}{(r+1)(r+2)(r+3)}##

Therefore ##\dfrac{3}{r(r+1)(r+2)(r+3)}## is of the form ##f(r)-f(r+1)##

When ##r=1## we have

##\left[\dfrac{1}{6}- \dfrac{1}{24}\right]##

When ##r=2## we have

##\left[\dfrac{1}{24}- \dfrac{1}{60}\right]##

When ##r=3## we have

##\left[\dfrac{1}{60}- \dfrac{1}{120}\right]##

...

When ##r=n## we shall have

##\dfrac{1}{n(n+1)(n+2)}- \dfrac{1}{(n+1)(n+2)(n+3)}##

Therefore;

##3\sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)(r+3)}=\left[\dfrac{1}{6}-\dfrac{1}{(n+1)(n+2)(n+3)}\right]##

##\sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)(r+3)}=\left[\dfrac{1}{18}-\dfrac{1}{3(n+1)(n+2)(n+3)}\right]##

Your insight welcome...refreshing on this today...