Question

Use mathematical inductions prove the following statements are true for all positive integers n.

Answer 1

For the first case, n=1,
\[\Large\bf \frac{1}{i(i+1)}=\frac{1}{1(1+1)}=\frac{1}{2},\]
\[\Large\bf \frac{n}{n+1}=\frac{1}{1+1}=\frac{1}{2}\]So it's true for the first case.

Answer 2

We'll assume it's true for the n\(\Large ^{th}\) case,
And try to show that it's true for the (n+1)\(\Large ^{th}\) case.

Answer 3

For the n\(\Large^{th}\) case we have,
\[\Large\bf \color{#DD4747}{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n(n+1)}}\quad=\quad \frac{n}{n+1}\]So we're assuming that's true.
Hopefully why I colored that left side will make sense in a moment.

Answer 4

So then for the (n+1)\(\Large ^{th}\) case,\[\large \color{#DD4747}{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n(n+1)}}+\frac{1}{(n+1)((n+1)+1)}\quad=\quad \frac{n+1}{(n+1)+1}\]

Answer 5

Grr it got cut off :(

Answer 6

So we know that the pink part is n/(n+1) based on what we assumed before.
So let's plug it in.\[\large \color{#DD4747}{\frac{n}{n+1}}+\frac{1}{(n+1)((n+1)+1)}\quad=\quad \frac{n+1}{(n+1)+1}\]

Answer 7

And then lemme remember what we were trying to do mmmmmm

Answer 8

Ok let's simplify down our denominators so they're easier to read,\[\Large\bf \frac{n}{n+1}+\frac{1}{(n+1)(n+2)}\quad=\quad \frac{n+1}{n+2}\]And from here we just need do a few steps to show that the right side is equivalent to the left side.

Answer 9

Getting a common denominator,\[\Large\bf \frac{n(n+2)+1}{(n+1)(n+2)}\quad=\quad \frac{n+1}{n+2}\]Expand out the numerator,\[\Large\bf \frac{n^2+2n+1}{(n+1)(n+2)}\quad=\quad \frac{n+1}{n+2}\]The numerator can factor,\[\Large\bf \frac{(n+1)(n+1)}{(n+1)(n+2)}\quad=\quad \frac{n+1}{n+2}\]We'll cancel out the common term,\[\Large\bf \frac{\cancel{(n+1)}(n+1)}{\cancel{(n+1)}(n+2)}\quad=\quad \frac{n+1}{n+2}\]Ok good so we end up with:\[\Large\bf \frac{n+1}{n+2}\quad=\quad \frac{n+1}{n+2}\]

Answer 10

And so we were able to show the equation is true for the (n+1)\(\Large ^{th}\) case.

So uhhhh yah that's kinda how that works, that whole induction thing.

Answer 11

thanks now i know how to solve :)