# Thread: Lost in proving .

1. ## Lost in proving .

If S is the sum of the series $\displaystyle 1+3x+5x^2+...+(2n+1)x^n$ , by considering (1-x)S , or otherwise , prove that .

$\displaystyle S=\frac{1+x-(2n+3)x^{n+1}+(2n+1)x^{n+2}}{(1-x)^2}$

where x is not 1 .

Thanks a lot .

2. Hi

I am gonna give you general proof for this AGP, Arithmetico-Geomtric Series

$\displaystyle a , (a+d) ,(a+2d)........$ is an AP and $\displaystyle b ,br,br^2.......$ is a GP

The series we have is
$\displaystyle ab ,(a+d)br,(a+2d)br^2,.......$

Now
$\displaystyle S_n=~ab +(a+d)br+(a+2d)br^2+.......+(a+(n-1)d)br^{n-1}$........(1)

$\displaystyle rS_n=~abr +(a+d)br^2+(a+2d)br^3+.......+(a+(n-1)d)br^{n}$..........(2)

We now substract (1) -(2)

$\displaystyle (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n$

$\displaystyle = \frac{ab}{1-r} + \frac{dbr(1-r^{n-1}}{(1-r)^2} - (a+(n-1)d)br^n$

$\displaystyle \implies ~S_n = ~ \frac{ab}{1-r} + \frac{dbr(1-r^{n-1})}{(1-r)^2} - \frac{(a+(n-1)d)br^n}{1-r}$
.................................................. ...............................................

Now you should always remember the way of proof of this thing for any general term and apply as it is in your case

For going further , when -1< r <1

$\displaystyle Lim_{n\rightarrow \infty}~S_n = \frac{ab}{1-r} + \frac{dbr}{(1-r)^2}$

Hi

Now
$\displaystyle S_n=~ab +(a+d)br+(a+2d)br^2+.......+(a+(n-1)d)br^{n-1}$........(1)

$\displaystyle rS_n=~abr +(a+d)br^2+(a+2d)br^3+.......+(a+(n-1)d)br^{n}$..........(2)

We now substract (1) -(2)

$\displaystyle (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n$
I can't understand this part . How did you get

$\displaystyle (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n$

when you subtract 2 from 1 .

Thanks again .

H
Now
$\displaystyle S_n=~ab +(a+d)br+(a+2d)br^2+.......+(a+(n-1)d)br^{n-1}$........(1)

$\displaystyle rS_n=~abr +(a+d)br^2+(a+2d)br^3+.......+(a+(n-1)d)br^{n}$..........(2)

We now substract (1) -(2)

$\displaystyle (1-r)S_n = ab+dbr + dbr^2 + ....dbr^{n-2} + dbr^{n-1} - (a+(n-1)d)br^n$

( 1) - (2) is given by

$\displaystyle (1-r)S_n = ab -abr +(a+d)br -abr^2+ (a+d)br^2 + ...... - (a+(n-1)d)br^n$

$\displaystyle (1-r)S_n = ab~ +(a+d)br -abr + (a+d)br^2 -abr^2+ ...... - (a+(n-1)d)br^n$

See the things with -ve sign are from (2)

Now open the brackets and go ahead to my step

Hi

$\displaystyle \implies ~S_n = ~ \frac{ab}{1-r} + \frac{dbr(1-r^{n-1})}{(1-r)^2} - \frac{(a+(n-1)d)br^n}{1-r}$
.................................................. ...............................................

THanks . If i plug these values in , then i will get my proof ?

a=1
b=1
r=x

6. I have just given you a general proof to it.
What you should do-

- Don't just plug the values you won't be able to remember the formula for long

-Try to remember the way it was proved and find the value of sum, for your a,b,r & d every time you get this type of question

-If you do it this way , every question of A.G.P. will be solved