Question

Prove by Induction:
\[\sum_{i=1}^{n} i/2^{i} =2-((n+2)/2^{n})\]
Prove by Induction:
\[\sum_{i=1}^{n} i/2^{i} =2-((n+2)/2^{n})\]
@Mathematics

Answer 1

I just can't seem to find the inductive conclusion

Answer 2

what you need to show is that
\[2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n+3}{2^{n+1}}\] right?i think

Answer 3

is that what you have?

Answer 4

yeah

Answer 5

so i guess it is algebra from here on in. i didn't do it, i just wrote it. let me see what i get

Answer 6

I just can't get it to simplify and equal out

Answer 7

yeah i get a numerator of
\[3n+5\]!!

Answer 8

that's what I get too

Answer 9

but if it was minus instead of plus it would give the right thing

Answer 10

hold on i think maybe this formula is not quite right

Answer 11

i am a moron! totally forgot arithmetic!

Answer 12

what is our error?

Answer 13

our error is a really annoying one. we both thought that 2 - 3 + 5 = 2 - 8 !

Answer 14

oh man

Answer 15

in other words we forgot that we are subtracting, not adding!

Answer 16

they are going to take away my math license now for sure. i bet it is clear now yes? you need
\[\frac{n+1}{2^{n+1}}-\frac{n+2}{2^n}\] and i bet it works out right.

Answer 17

yes hehe don't forget to carry out the minus sign!!

Answer 18

wait hold on, what about the 2?

Answer 19

it works fine. you want
\[2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n+3}{2^{n+1}}\] the 2 in front just stays there.

Answer 20

ahhh I see lol

Answer 21

just make sure to compute the second part correctly. it is
\[\frac{n+1}{2^{n+1}}-\frac{n+2}{2^n}\]
\[\frac{n+1-2(n-2)}{2^{n+1}}\]
\[\frac{-n-3}{2^{n+1}}\]
\[-\frac{n+3}{2^{n+1}}\]
just as you want it

Answer 22

typo there , line 2 should be
\[\frac{n+1-2(n+2)}{2^{n+1}}\]

Answer 23

yeah, I've got it thanks
This problem just tricks you into forgetting simple arithmetic
Couldn't figure out what was going wrong for a while lol

Answer 24

the old "order of operations" strikes again. please excuse my senile old aunt