Prove
$$ \sum_{n=1}^\infty \frac { n(n-1)}{2^n} =4 $$

Hint: Use $$ \sum_{n=0}^\infty x^n =\frac 1 {1-x}, \quad |x|

Question

Prove
$$ \sum_{n=1}^\infty \frac { n(n-1)}{2^n} =4 $$

Hint: Use  $$ \sum_{n=0}^\infty  x^n =\frac 1 {1-x}, \quad |x|<1 $$

and use the term by term differentiation method.
Of course, I know how to do it. I am sure some of you can. Give it a try.

anonymous · Answer

The problem needs differentiation twice. 
This one needs three:


$$ \sum_{n=3}^\infty \frac { n(n-1)(n-2)}{2^n} =12 $$

anonymous · Answer

Why differentiation? It isn't continuous is it?

anonymous · Answer

for |x| < 1 it is continuous and differentiable.

anonymous · Answer

Hint: Use  $$ \sum_{n=0}^\infty  x^n =\frac 1 {1-x}, \quad |x|<1 $$ 
Differentiate both sides to get
  $$ \sum_{n=1}^\infty  n x^{n-1} =\frac {-1} {(1-x)^2}, \quad |x|<1 $$ 
Differentiate another time and multiply both sides by $$x^2$$
and you can find the first sum.

experimentx · Answer

one question: why change from n=0 to n=1 on differentiation??

anonymous · Answer

Yeah, I should have considered your hint earlier :(

anonymous · Answer

1st is Zero anyway

anonymous · Answer

I mean for n=0

experimentx · Answer

Oo.. i get it.

anonymous · Answer

Yes, you did.

experimentx · Answer

for x=(1/2) we get, 2*(x)^2/(1-x)^3 = 4 , on the left hand side
summation n=2 to inf n(n-1)(1/2)^2 = 4

experimentx · Answer

Problem was half solved by you anyway :D

anonymous · Answer

@experimentX do the problem that involves 3 differentiations

experimentx · Answer

i'll give a try

anonymous · Answer

You are almost there.

experimentx · Answer

yeah i get it,

multiply both sides by x^3 and replace x=1/2, we get the required expression on left hand side, on the right hand side, we have 
$$\frac{6x^3}{(1-x)^4} = \frac{6*(1/2)^3}{(1/2)^4} = 12 $$

anonymous · Answer

Excellent. What is your background in Math?

experimentx · Answer

Physics major

anonymous · Answer

From my experience, Physics majors are very good in Math,

experimentx · Answer

I am quite not ... still struggling with calculus, though i appeared math major with algebra,  analysis, advanced calculus, and mechanics ... i never understood anything .. esp algebra and analysis.

anonymous · Answer

You will get there. Practice, Practice and Practice.

experimentx · Answer

therefore still struggling on physics ... at least i hope to finish differential equations and multivariable calculus ...

anonymous · Answer

Use also my site http://saab.org to practice math from algebra to calculus to actuarial problems.

anonymous · Answer

I will post soon another seires problem. Try to do it. I will close this problem for now.

experimentx · Answer

thanks .. i will. you don't look like a student ... are you teacher at some university?