Question

Mathematical induction problem

Answer 1

Prove that for any positive integer n, \[\sum_{k=1}^{n} k2^{k} = (n-1)2^{n+1}+2\]

Answer 2

start induction proving this is true for n=1

Answer 3

\[\sum_{k=1}^{1}1\times2^{1}=2\]
\[(1-1)2^{1+1}+2=2\]

How do I do the inductive step?

Answer 4

good

Answer 5

now suppose that, for m>1
\[\large \sum_{k=1}^mk2^k=(m-1)2^{m+1}+2 \]
is true. This is your induction hypothesis.
Get it?

Answer 6

it is basically what u had with m instead of n

Answer 7

now instead of m state the problem with m+1 instead of n

Answer 8

so after proving for n=1, we assume it works for all n?

Answer 9

no. we assume it works for m>1 but we have to it works for m+1

Answer 10

\[\sum_{k=1}^{m+1}k2^{k}= \sum_{k=1}^{1}k2^{k} + \sum_{k=1}^{m}k2^{k}\]

Answer 11

go ahead, now replace the induction hypothesis

Answer 12

\[2 + \sum_{k=1}^{m}k2^{k} = (m)2^{m+2} + 2\]

Answer 13

yes, but it would be better if you show all your work (every step) when doing this kind of problems

Answer 14

what would I do with this? has this proved the inductive step?
\[\sum_{k=1}^{m}k2^{k}=(m)2^{m+2} \]

Answer 15

the induction thesis was
\[\large \sum_{k=1}^{m+1}k2^k=m2^{m+2}+2 \]

Answer 16

this is what you have to prove. and that is what u did.

Answer 17

ah alright i see thank you

Answer 18

u r welcome