Question

please prove this by induction for all non-negative integers n.
1/2 + 2/(2^2) +3/(2^3) +...+n/(2^n) = 2- (n+2/(2^n))

Answer 1

start by proving it true for n = 1

1st term is 1/2 now substitute n = 1 into the right hand side of the equation.

Answer 2

and show the sum of 1 term is 1/2

Answer 3

that is the 1st step to induction

Answer 4

the next step is to assume its true for a value n = k

so the sum of k terms is

\[2 - (\frac{k + 2}{2^k})\]

Answer 5

thanks. what about the last step n=k+1 ?

Answer 6

yes so find the k + 1 term and add it to the sum of k terms.

then show it is equal to the sum of k + 1 terms.

Answer 7

so you need to show

k +1 term is

\[\frac{k + 1}{2^{k + 1}}\]

then show

\[S_{k} + t_{k + 1} = S_{k + 1}\]

Answer 8

oops should read

\[2 = \frac{k + 2}{2^k} + \frac{k + 1}{2^{k + 1}} = 2 - \frac{(k + 1) + 2}{2^{k + 1}}\]

Answer 9

dang... it should read 2 - ... not 2 =

Answer 10

thanks a lot. really helping

Answer 11

its reasonably easy from here... just be careful with the signs of terms...

and use the fact that

\[2^k = \frac{2}{2^{k + 1}}\]