Question

Show by complete Induction \(\forall n \in \mathbb{N}\):
\[\large\sum_{k=1}^{n}kx^{k} = \frac{nx^{n+1}}{x-1}-\frac{x^{n+1}-x}{(x-1)^{2}}\]

Answer 1

I know i need to start firs with setting n=0 and n=n+1 but i dont get any meaningfull solution

Answer 2

Try googling for Arithmetico Geometric Progressions

Answer 3

I think you need k's where you have n's
can I edit your Q again?

Answer 4

yes naturally TuringTest..

Answer 5

is that right, or were some of those actually supposed to be n's ?

Answer 6

you are right TuringTest i have one k wrong i edit it wait

Answer 7

now its ok, if you refresh you have right version..

Answer 8

am I just messing up or am I getting this as not true for n=1

Answer 9

I have the complete file in attachment you can take a look

Answer 10

@TuringTest i understand what u mean right now, i also didnt get a meaningfull solution for n=1 but i am noob at mathematics.. :) maybe i do wrong..

Answer 11

no it's correct for n=1 I was mistaken

Answer 12

ok :)

Answer 13

@TuringTest can you post your solution pls ?

Answer 14

I don't have a proof, but you want me to show that it is true for n=1 ?

Answer 15

Proposition : \(\Large \sum_{k=1}^{n}kx^{k} = \frac{nx^{n+1}}{x-1} - \frac{x^{n+1}-x}{(x-1)^{2}}\)

Proposition for n= 1: \(\Large LHS = x, RHS = \frac{1.x^{2}}{x-1}-\frac{x^{2}-x}{(x-1)^{2}}\)
\(\Large RHS = \frac{x^{2}}{x-1}-\frac{x(x-1)}{(x-1)^{2}} = \frac{x^{2}-x}{x-1}=x\)
Thus the proposition is true for n = 1.

Answer 16

(I mean I have no complete proof by induction yet)

Answer 17

yeah, FoolAround beat me to it....

Answer 18

now, let us assume the proposition is true for \(n=m\). or
\[\Large \sum_{k=0}^{m} = \frac{mx^{m+1}}{x-1}- \frac{x^{m+1}-x}{(x-1)^{2}}\]
We have to show that the proposition is true for \(n = m+1\) given that the proposition is true for \(n=m\)

Answer 19

ok..

Answer 20

@FoolAroundMath thank you but i couldnt understand how is it possible ?\[\frac{x^{2}-x}{x-1}=x\]

Answer 21

factor out x on the top

Answer 22

uggh at the typos:
in prev post it should be k=1 to k=m. Rewriting proof now >_<

Answer 23

you should always copy before posting for that reason ;)
you clearly were going in th right direction, stripping out the last term

Answer 24

and the fact that OS keeps crashing does not help. Writing in notepad and copy pasting now.

Answer 25

So, we need to show that
\[\Large \sum_{k=1}^{m+1}kx^{k} = \frac{(m+1)x^{m+2}}{x-1} - \frac{x^{m+2}-x}{(x-1)^{2}}\]
So, lets start from the left hand side of the equation.
\[\Large \sum_{k=1}^{m+1}kx^{k} = (m+1)x^{m+1}+\sum_{k=1}^{m}kx^{k}\]Now we know the value of the sum from k=1 to k=m. So substituting that, we get
\[\Large \frac{(m+1)x^{m+1}(x-1)}{x-1}+\frac{mx^{m+1}}{x-1} - \frac{x^{m+1}-x}{(x-1)^{2}}\] I've multiplied (x-1) with both numerator and denominator for \((m+1)x^{m+1}\)

Answer 26

\[\Large \frac{(m+1)x^{m+2}}{x-1} - \frac{(m+1)x^{m+1}}{x-1}+\frac{mx^{m+1}}{x-1}-\frac{x^{m+1}-x}{(x-1)^{2}}\]
\[\Large \frac{(m+1)x^{m+2}}{x-1} - \frac{mx^{m+1}}{x-1}-\frac{x^{m+1}}{x-1}+\frac{mx^{m+1}}{x-1}-\frac{x^{m+1}-x}{(x-1)^{2}}\]
\[\Large \frac{(m+1)x^{m+2}}{x-1} - \frac{x^{m+1}(x-1)}{(x-1)^{2}}-\frac{x^{m+1}-x}{(x-1)^{2}}\]
\[\Large \frac{(m+1)x^{m+2}}{x-1} - \frac{x^{m+2}}{(x-1)^{2}}+\frac{x^{m+1}}{(x-1)^{2}}-\frac{x^{m+1}-x}{(x-1)^{2}}\]
\[\Large \frac{(m+1)x^{m+2}}{x-1} - \frac{x^{m+2}}{(x-1)^{2}}+\frac{x}{(x-1)^{2}}\]
\[\Large \frac{(m+1)x^{m+2}}{x-1} - \frac{x^{m+2}-x}{(x-1)^{2}}\]which is what we had to show

Answer 27

\[\frac{x^{2}-x}{x-1}=x\] how is it possible ?

Answer 28

\(x^{2}-x = x(x-1) \)

Answer 29

\[\frac{x^2-x}{x-1}=\frac{x\cancel{(x-1)}}{\cancel{x-1}}=x\]

Answer 30

aah ok thanks @FoolAroundMath and @TuringTest

Answer 31

@TuringTest what is the LaTeX for strikeout ?

Answer 32

@FoolAroundMath i think you mean this, its with cancel
\frac{x\cancel{(x-1)}}{\cancel{x-1}}=x