Induction. Prove that n=k+1 on a rather tricky example (for me). Help needed?

This has been bugging me for 2 days now, I can't seem to find the solution. I've multiplied the first fraction by (k+2) and that's as far as I could go. Here is the problem: enter image source here
It would be more than helpful if you could explain how to solve this. Thanks

1 Answer
Monzur R.
May 22, 2018

See below

Explanation:

I assume you want to prove

#sum_(k=1)^n(n^2+n+1)/(n(n+1))=(n(n+2))/(n+1)#

Let #P(n)# be the statement

#sum_(k=1)^n(k^2+k+1)/(k(k+1))=(n(n+2))/(n+1)#

for #ninNN#. We will prove #P(n)# holds for all #ninNN# using induction.

First, we check #P(1)#.

#sum_(k=1)^1(k^2+k+1)/(k(k+1))=3/(1(2))=3/2=(1(1+2))/(1+1)# so #P(1)# holds

Now we assume that #P(n)# is true for some #ninNN#. That is, that

#sum_(k=1)^n(k^2+k+1)/(k(k+1))=(n(n+2))/(n+1)#

for some #ninNN#.

Then

#sum_(k=1)^(n+1)(k^2+k+1)/(k(k+1))=sum_(k=1)^n(k^2+k+1)/(k(k+1))+((n+1)^2+n+1+1)/((n+1)(n+2))#

#=(n(n+2))/(n+1)+(n+1)/(n+2)+1/(n+1)#

#=(n^2+2n+1)/(n+1)+(n+1)/(n+2)#

#=(n+1)^2/(n+1)+(n+1)/(n+2)#

#=((n+1)(n+2)+(n+1))/(n+2)#

#=((n+1)((n+2)+1))/(n+2)#

#=((n+1)((n+1)+2))/((n+1)+1)#

so #P(n+1)# holds.

In particular, #P(n)rArrP(n+1)# and #P(1)# holds so, by the induction axiom on #NN#, #P(n)# holds for all #ninNN#.

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