# Using the principle of Mathematical Induction, prove that `1 / (1*2*3) + 1 /(2*3*4) +.....+1 / (n(n+1)(n+2)) = 1/4-1 /(2(n+1)) +1/(2(n+2))` For every positive integer n.

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When n = 1

`LHS`

`= 1/(1*2*3)`

`= 1/6`

`RHS`

`= 1/4-1/(2(1+1))+1/(2(1+2))`

`= 1/4-1/4+1/6`

`= 1/6`

when n = 1 the result is true.

Let us assume when n = p where p is a positive integer the result is true.

`1/(1xx2xx3)+.......1/(p(p+1)(p+2)) = 1/4-1/(2(p+1))+1/(2(p+2))`

When n = p+1

We have to prove that ;

`RHS = 1/4-1/(2(p+1+1))+1/(2(p+1+2))`

`RHS = 1/4-1/(2(p+2))+1/(2(p+3))`

We can get the sum of (p+1) terms when we add the (p+1) term to the sum of p terms at RHS.

RHS

`= 1/4-1/(2(p+1))+1/(2(p+2))+1/((p+1)(p+2)(p+3))`

Using partial fractions

`1/((p+1)(p+2)(p+3)) = A/(p+1)+B/(p+2)+C/(p+3)`

`1 = A(p+2)(p+3)+B(p+1)(p+3)+C(p+1)(p+2)`

When p = -1 we get A = 1/2

When p = -2 then we get B = -1

When p = -3 then we get C = 1/2

RHS

`= 1/4-1/(2(p+1))+1/(2(p+2))+1/((p+1)(p+2)(p+3))`

`= 1/4-1/(2(p+1))+1/(2(p+2))+1/(2(p+1))-1/((p+2))+1/(2(p+3))`

`= 1/4-1/(2(p+2))+1/(2(p+3))`

This is the answer that we need for n = p+1

So for n = p+1 the result is true.

**So from mathematical induction the result is true.**

`p(n):1/(1.2.3)+1/(2.3.4)+................+1/(n.(n+1).(n+2))=1/4-1/(2(n+1))+1/(2(n+2))`

`n=1`

`LHS=1/(1.2.3)=1/6`

`RHS=1/4-1/4+1/(2.3)=1/6`

Thus P(1) is true.

Let us assume P(k) is true.

`P(k):1/(1.2.3)+1/(2.3.4)+................+1/(k.(k+1).(k+2))=1/4-1/(2(k+1))+1/(2(k+2)) `

`` Now we wish to prove P(k+1) is true when P(k) is true.i.e

`P(k+1):1/(1.2.3)+1/(2.3.4)+................+1/(k.(k+1).(k+2))+1/((k+1)(k+2)(k+3))=1/4-1/(2(k+2))+1/(2(k+3))`

`` `LHS=1/(1.2.3)+1/(2.3.4)+................+1/(k.(k+1).(k+2))+1/((k+1)(k+2)(k+3))`

`=1/4-1/(2(k+1))+1/(2(k+2))+1/((k+1)(k+2)(k+3))`

`=1/4-((k+2)(k+3)-(k+1)(k+3)-2)/(2(k+1)(k+2)(k+3))`

`=1/4-(k^2+5k+6-k^2-4k-3-2)/(2(k+1)(k+2)(k+3))`

`=1/4-(k+1)/(2(k+1)(k+2)(k+3))`

`=1/4-1/(2(k+2)(k+3))`

`RHS=1/4-1/(2(k+2))+1/(2(k+3))`

`=1/4-(k+3-k-2)/(2(k+2)(k+3))`

`=1/4-1/(2(k+2)(k+3))`

`=LHS`

`Thus`

P(k+1) is true when P(k) is true. So by principle of mathenmatical induction P(n) is true for all n.