David H. answered • 01/23/17

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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 k

^{3}+6k^{2}+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

David H.

No problem! Feel free to message me if you have any other questions

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01/23/17

Emmanuel C.

_{thank u very}much01/23/17