1 1 ⋅ 2 ⋅ 3 \frac{1}{1\cdot2\cdot3}

Algebra Level 3

1 1 ⋅ 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 4 + 1 3 ⋅ 4 ⋅ 5 + ⋯ \large \dfrac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4\cdot 5}+\cdots

If the series above can be expressed as p q \dfrac pq , where p p and q q are coprime positive integers , find p + q p+q .


The answer is 5.

View wiki
Math Man by math man , 20, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sumanth R Hegde
Feb 13, 2017

The summation is S = ∑ r = 1 ∞ T r \displaystyle S = \sum_{r=1}^\infty {T_r} where T r = 1 r ( r + 1 ) ( r + 2 ) = 1 2 r + 2 − r r ( r + 1 ) ( r + 2 ) = 1 2 ( 1 r ( r + 1 ) − 1 ( r + 1 ) ( r + 2 ) ) \begin{aligned} \displaystyle T_r = \frac{1}{r(r + 1)(r +2)} = \frac{1}{2} \frac{ r +2 -r }{r(r+1)(r+2)} = \frac{1}{2}( \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} ) \end{aligned}

Thus S is a telescopic sum .We end up with

S = 1 4 \boxed{\color{#D61F06}{S = \dfrac{1}{4} }}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...