Question

induction step for 1*2^1+2*2^2+3*2^3+...+k*2^k=(k-1)2^(k-1)+2?

Answer 1

I plugged k-1 in already and got
\[(k-1)2^{k-1} + 2 + (k+1) * 2^{k+1}=k*2^k+2\]

Answer 2

I just can't seem to simplify the left side to make it the same as the right

Answer 3

\[(k*2^k)/2 -((5*2^k)/2)+1\] this is what I ended up on the left side

Answer 4

1st show its true for k=1
then assume its true for, say, k=j
then prove its also true for k=j+1

Answer 5

I dont think it is true for k=2
(2-1)2^(2-1) +2=1*2+2=4 not right..

Answer 6

you are correct it is not true

Answer 7

than induction wont work

Answer 8

yeah I wrote it wrong it's suppose to be (k-1)2^(k+1)+2

Answer 9

1*2^1+2*2^2+3*2^3+...+k*2^k=(k-1)2^(k+1)+2 but I still can't get it thought I could

Answer 10

give me a sec I will write it down.

Answer 11

so plugging in k+1 I get
\[(k-1)*2^{k+1}+2 +(k+1)*2^{k+2}=k*2^k+2\]

Answer 12

oops the right part is k*2^(k+2)+2

Answer 13

yes and thats the end of induction. You showed that for k+1 that formula stands.

Answer 14

no because I couldn't show that the left side is equal to the right..

Answer 15

how do I simply so that it works?

Answer 16

simplify*

Answer 17

1*2^1+2*2^2+3*2^3+...+k*2^k=(k-1)2^(k+1)+2

so for k+1 we need that the result should be (k+1-1)2^(k+1+1) +2
and this is what you did

Answer 18

but how about the left side? the only reason why I got (k)2^(k+2)+2 because I plugged in k+1 on the right side but don't you need to also prove that the left side equals the right if you get what I'm trying to say