Prove by Mathematical induction

[Erros]

I can't do this task properly, can I get help and instructions, I've tried but I'm not doing well :/

 

[topsquark]

The reason you are having a problem with it is that it isn't true! Your base case n = 1 gives [math]\dfrac{1}{1 \cdot 2} \neq \dfrac{2 \cdot 1}{2 \cdot 1 + 1}[/math]
Are you sure you copied the problem correctly?

-Dan

 

[BigBeachBanana]

I can't do this task properly, can I get help and instructions, I've tried but I'm not doing well :/View attachment 32638

I believe the problem statement written is wrong that's why you're having trouble establishing the base case.
For n=1, the LHS is: \(\displaystyle \frac{1}{(2\times 1)(2\times1+1)}=\frac{1}{2(3)}\). This doesn't fit the first term \(\displaystyle \frac{1}{1(2)}\).

Secondly, for n=1 the RHS is:\(\displaystyle \frac{2(1)}{2(1)+1}=\frac{2}{3}\). This is actually the result of the sum of first 2 terms i.e. \(\displaystyle \frac{1}{2}+\frac{1}{2(3)}=\frac{2}{3}\)

 

[lex]

The problem is [imath]\sum \limits_{r=1}^{2n}\tfrac{1}{r(r+1)} = \tfrac{2n}{2n+1}[/imath]

 

[Erros]

This is an update, can anyone help me do more for n = k + 1, that part is not clear to me yet, I did these 2 steps, I still miss n = k + 1 @BigBeachBanana @topsquark

 

[BigBeachBanana]

The problem is [imath]\sum \limits_{r=1}^{2n}\tfrac{1}{r(r+1)} = \tfrac{2n}{2n+1}[/imath]

This is an update, can anyone help me do more for n = k + 1, that part is not clear to me yet, I did these 2 steps, I still miss n = k + 1 @BigBeachBanana @topsquark

If I were to do this problem, I would simplify the problem by letting [imath]m=2n[/imath]. Then the problem becomes
[math]\sum \limits_{r=1}^{m}\tfrac{1}{r(r+1)} = \tfrac{m}{m+1}[/math]This is easier for you to establish the base case ([imath]m=1[/imath]) and the [imath]m+1[/imath] case which is equivalent of [imath]2n+1[/imath].

 

[Steven G]

The nth term is NOT in the form \(\displaystyle \dfrac {1}{(2n)(2n+1)}.\)

Note that 2n<2n+1. The smaller number in each denominator is not always even, while 2n is.

The last term is \(\displaystyle \dfrac {1}{(2n)(2n+1)}.\) All that means is that the last term can be 1/(8*9) but not 1/(9*10). The smaller number in the last denominator must be even.

The nth term is in the form \(\displaystyle \dfrac {1}{(n)(n+1)}.\)