Question

use mathematical induction to prove

n sigma i=1 1/i(i+1) = n/n+1

Answer 1

posting a solution in a sec, how familiar are you with proofs by induction?

Answer 2

slightly familiar

i know 3 steps

i think im just having trouble moving forward from

k sigma i = 1 1/i(i+1) + 1/ (k+1)(k+2)

Answer 3

alright, so you added the "missing term"
\[\frac{1}{(k+1)(k+2)}\]
to your summation, and you want to simplify? just trying to make sure we are the same page since there are a couple of ways to tackle this problem.

Answer 4

yes, i want to simplify

Answer 5

by the way, this is my solution, but I would still like to do it with your approach.

Answer 6

alright so you have:
\[\frac{1}{(k+1)(k+2)}+\sum_{i=1}^{k}\frac{1}{i(i+1)} \]
to simply, use your inductive hypothesis, which is that the summation equals k/(k+1). Then we have:
\[\frac{1}{(k+1)(k+2)}+\frac{k}{k+1}\]
and the game is to simplify as much as we can.
i believe that work is done in my 3rd pic.

Answer 7

but how do i know the summation is k/k+1 and not 1/k+1?

Answer 8

because we are trying to prove:
\[\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}\]
We assume this is true for some fixed natural number 'k'. so we have:
\[\sum_{i=1}^{k}\frac{1}{i(i+1)} = \frac{k}{k+1}\]
that is why we use k/(k+1), not 1/(k+1)

Answer 9

gotta use the formula you are trying to prove.

Answer 10

oh... right.

Answer 11

you also have to know what you are aiming for as well. hopefully, those fractions should simplify to:
\[\sum_{i=1}^{k+1}\frac{1}{i(i+1)}= \frac{(k+1)}{(k+1)+1} = \frac{k+1}{k+2}\]

Answer 12

ok now im seeing it...

Answer 13

sweet :) induction is a powerful tool, but getting the hang of it can be rough sometimes (especially with inequalities, they can die in a fire >.<)

Answer 14

so 1/k+1 + k/k+1 = k+1/K+1?

Answer 15

that last kkay is not meant to be a capital

Answer 16

you want to show that:
\[\frac{1}{(k+1)(k+2)} + \frac{k}{k+1} \Rightarrow \frac{k+1}{k+2}\]
somehow, by combining fractions, simplifying, etc.

Answer 17

yeah, im just looking at combining the denominator... because on the numerator, i can see that its just 1+k... but on the bottom does (k+1)(k+2) + k+1 just combine to (k+1)(k+2) ?

Answer 18

you gotta get a common denominator before you add those fractions up:
\[\frac{1}{(k+1)(k+2)}+\frac{k}{k+1} = \frac{1}{(k+1)(k+2)} +\frac{k(k+2)}{(k+1)(k+2)}\]
\[= \frac{1+k(k+2)}{(k+1)(k+2)}\]

Answer 19

oh damn

Answer 20

the rest of the solution is in the 3rd pic (except i used the dummy variable n, instead of k)

Answer 21

ah eys
now on to the next induction :)