# Use the principles of mathematical induction to prove `1/(1*2*3) + 1/(2*3*4) +1/(3*4*5)+...+1/(n(n+1)(n+2)) = (n*(n+3))/(4(n+1)(n+2))`

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### 1 Answer

The relation `1/(1*2*3) + 1/(2*3*4) +1/(3*4*5)+...+1/(n(n+1)(n+2)) = (n*(n+3))/(4(n+1)(n+2))` has to be proved. This can be done using mathematical induction.

First take the base case n = 1: the right hand side is `1/(1*2*3) = 1/6` and the left hand side is also `(1*4)/(4*2*3) = 1/6`

Assume the relation holds for an integer n

=> `1/(1*2*3) + 1/(2*3*4) +1/(3*4*5)+...+1/(n(n+1)(n+2)) = (n*(n+3))/(4(n+1)(n+2))`

For the case of the integer n + 1,

`1/(1*2*3) + 1/(2*3*4) +1/(3*4*5)+...+1/(n(n+1)(n+2)) + 1/((n+1)(n+2)(n+3))`

=> `(n*(n+3))/(4(n+1)(n+2)) + 1/((n+1)(n+2)(n+3))`

=> `(n*(n+3)(n+3) + 4)/(4*(n+1)(n+2)(n+3))`

=> `(n*(n^2 + 6n + 9) + 4)/(4*(n+1)(n+2)(n+3))`

=> `(n^3 + 6n^2 + 9n + 4)/(4*(n+1)(n+2)(n+3))`

=> `((n+1)^2*(n+4))/(4*(n+1)(n+2)(n+3))`

=> `((n+1)(n+4))/(4*(n+2)(n+3))`

**This proves the relation `1/(1*2*3) + 1/(2*3*4) +1/(3*4*5)+...+1/(n(n+1)(n+2)) = (n*(n+3))/(4(n+1)(n+2))` is true for all integral values.**