Originally Posted by

**Lprdgecko** I apologize in advance for not knowing how to type mathematical notation, I'll do the best I can to make the question as clear as possible.

The problem is:

Show that (sigma)(i=1 to n) i/(2^i) = 2 - (n+2) / (2^n) for all natural numbers n.

Here is what I did:

Proof. We proceed by weak mathematical induction.

Base Case. Put n = 1. Then (sigma)(i=1 to n) i/(2^i) = (sigma)(i=1 to 1) i/(2^i) = 1/(2^1) = 1/2 = 2 - (1+2)/(2^1) = 2 - (n+2)/(2^n) , so the base case is established.

Induction Step. Fix a natural number n and suppose (sigma)(i=1 to n) i/(2^i) = 2 - (n+2)/(2^n).

Observe that (sigma)(i=1 to (n+1)) i/(2^i)

= 1/(2^1) + 2/(2^2) + 3/(2^3) +...+ (n+1)/(2^(n+1))

= [1/(2^1) + 2/(2^2) + ... + n/(2^n)] + (n+1)/(2^(n+1))

= [(sigma)(i=1 to n) i/(2^i)] + (n+1)/(2^(n+1))

= [2 - (n+2)/(2^n)] + (n+1)/(2^(n+1))

= [2 - (2)(n+2) / (2)(2^n)] + (n+1)/(2^(n+1))

= [2 - (2n+4) / (2^(n+1))] + (n+1)/(2^(n+1))

= 2 - (3n+5) / (2^(n+1))

= 2 - (3n + 3 + 2) / (2^(n+1))

= 2 - [3(n+1) + 2] / (2^(n+1))

My goal was to show 2 - [(n+1)+2] / (2^(n+1)) ... I got really close except for the extra 3. Did I do something wrong, or is the proof finished?

Again, I apologize if this is difficult to read...