Question

What is this? A room full of answerers and no questioners to stimulate us?

Answer 1

\[Prove that ........ \sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 2^{k}=3^{n}\]

Answer 2

hahahah ! :P

Answer 3

I thought that was already proved ;)

Answer 4

proof by contradiction perhaps?

Answer 5

ill go find one or two

Answer 6

is that a fraction notation; n/k ?

Answer 7

or is it the nCr ?

Answer 8

Doesn't that have to be proved by induction?

Answer 9

nCr think binomial expansion theorem

Answer 10

with any luck well be going over proofs soon enough in discrete

Answer 11

maybe it is the friday of the holiday weekend!

Answer 12

this is from an old assignment in my combinatorics class

Answer 13

it works - does it matter why ?

Answer 14

id prolly start it with n=1 then n=2 then so on and so forth, simply becasue I like the feel of lead in my teeth

Answer 15

can you do it by induction?

Answer 16

proofs are one of my favourite parts in math, it does matter :)

Answer 17

I have one

Answer 18

share!

Answer 19

I tried induction, but I think something is wrong..

Answer 20

Oh, I see what I did.

Answer 21

Hrmph.. Still stuck.

Answer 22

trivial..binomial theorem

Answer 23

\[(a+b)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) a^kb^{n-k}\]

Answer 24

let a=2 and b=1

Answer 25

I'm thinking I must be on the wrong track here, but this is all I can come up with..
\[p(n): \sum_{k=0}^n {n\choose k}2^k = 3^n\]
\[p(0): \sum_{k=0}^0 {n\choose k}2^k = {0 \choose 0}2^0 = 1 = 3^0\]
\begin{align*}
p(n+1):& \sum_{k=0}^{n+1} {n+1\choose k}2^k \\
=& \frac{(n+1)!}{(n+1)!\;0!}2^{n+1} + \sum_{k=0}^n {n+1\choose k}2^k \\
=& 2^{n+1} + \sum_{k=0}^n \frac{(n+1) \cdot n!}{(n+1-k)\cdot k! (n-k)!}2^k \\
=& 2^{n+1} + (n+1)\sum_{k=0}^n \frac{1}{n+1-k}{n\choose k}2^k \\
\end{align*}

Answer 26

But I couldn't do anything with that -k in the denominator.

Answer 27

If you wanted to prove by induction then you might want to use ..

\[\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}\]