1/n(n+1)(n+2)

Question

_{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)}

Answer 1

2 Steps to mathematical induction.

 

1.) initial step. Show true for n = 1

1/(1*(1+1)(1+2)) = 1/6

1*(1+3)/(4*(1+1)(1+2)) = 1*4/(4*2*3) = 1/6

 

So holds.

2.) Now assume that the formula holds for some k > 1 where k is a natural number.

Need to show true for k+1. {i.e. we want to show 1/(1*2*3)+1/(2*3*4)+...+1/(k*(k+1)*(k+2))+1/((k+1)(k+2)(k+3)) = (k+1)*(k+4)/(4*(k+2)(k+3))  }

 

Thus by hypothesis

1/(1*2*3) + 1/(2*3*4) + .... + 1/(k*(k+1)*(k+2)) = k*(k+3)/(4*(k+1)(k+2))

Add to both sides 1/((k+1)(k+2)(k+3)).

The left side will be of the form we want.

 

Looking at right side you have

k*(k+3)/(4*(k+1)(k+2)) + 1/((k+1)(k+2)(k+3)) =

[k*(k+3)*(k+1)*(k+2)*(k+3)+4*(k+1)*(k+2)]/ [4*(k+1)*(k+2)*(k+1)*(k+2)*(k+3)]

 

You see a (k+1) and (k+2) cancel from numerator and denominator so you have

[k*(k+3)(k+3)+4]/[4*(k+1)*(k+2)*(k+3)]

 

The numerator becomes k³+6k²+9k+4. This can be factored out to (k+1)(k+4)(k+1)

Thus one of the (k+1) cancel from numerator and denominator and we are left with

(k+1)(k+4)/[4*(k+2)(k+3)]. This is exactly what we want on the right side.

 

Thus our formula holds for k+1.

 

Since k was arbitrary this means it holds for all n.

 

Hope helpful.

Best - David