Combinatorics Proof: Prove n5^n = \frac{5}{4} Sum

[pupeye11]

Homework Statement

Prove that

$$n5^n = \frac{5}{4} \sum_{k=0}^{n}k\begin{pmatrix}n\\k\end{pmatrix}4^k$$

(Hint: First Expand $$(1+x^2)^n$$)

The Attempt at a Solution

So if I expand that I get

$$(1+x^2)^n = (1+x^2)(1+x^2)...(1+x^2)$$ n times so it equals

$$\sum_{k=0}^n (1+x^2)$$

Not sure where to go from there, unless I expanded that wrong?

 

[Office_Shredder]

$$\sum_{k=0}^n (1+x^2)$$

Not sure where to go from there, unless I expanded that wrong?

This is adding (1+x²) a bunch of times, not multiplying it. Try using the binomial theorem to get the correct expansion

 

[njama]

Hello!

You need to use the binomial expansion formula for $$(1+x^2)^n$$ for x=2.

 

[pupeye11]

Ok, so I exapnded that using binomial expansion and evaluated it at x=2. I'll show my steps below

$$(1+x^2)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}(x^2)^k$$

$$(1+4)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}4^k$$

$$5^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}4^k$$

I believe that i then need to set $$5^n =4^k$$ and solve because the rest are constants I believe. I am drawing a major blank on that though...

 

[njama]

What are you talking about?
Hint:
Do something with:
$$<br /> k\begin{pmatrix}n\\k\end{pmatrix}<br />$$
like some algebraic stuff.

 

[pupeye11]

What? The binomial Theorem states that

$$(x+y)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}x^{n-k}y^k$$

so using that I got that

$$(1+x^2)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}(x^2)^k$$

Then evaluated that for x=2 to get

$$5^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}4^k$$

From there there is no k times n choose k... So I am not really sure where that came from...

 

[Dick]

What? The binomial Theorem states that

$$(x+y)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}x^{n-k}y^k$$

so using that I got that

$$(1+x^2)^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}(x^2)^k$$

Then evaluated that for x=2 to get

$$5^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix}4^k$$

From there there is no k times n choose k... So I am not really sure where that came from...

That's fine. But it's not the expression you want to prove. Try taking d/dx of both sides before you put x=2.

 

[pupeye11]

So if I take d/dx I'll get

$$n(1+x^2)^{n-1}(2x) = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k) (x^{2k-1})$$

which after putting 2 in for x I will then get

$$4n5^{n-1} = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k)(4^{2k-1})$$

Then from there I figured I could multiply both sides by 5^1 to get

$$4n5^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k)(4^{2k-1})(5)$$

but I am stuck from there.

 

[Dick]

So if I take d/dx I'll get

$$n(1+x^2)^{n-1}(2x) = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k) (x^{2k-1})$$

which after putting 2 in for x I will then get

$$4n5^{n-1} = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k)(4^{2k-1})$$

Then from there I figured I could multiply both sides by 5^1 to get

$$4n5^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k)(4^{2k-1})(5)$$

but I am stuck from there.

x^(2k-1) becomes 2^(2k-1) not 4^(2k-1). Clean that up.

 

[pupeye11]

Ok I fixed that which then I realized that

$$2^{2k-1} = 4^{k-1}$$

which I can then multiply both sides by $$4^{1}$$ to get

$$16n5^n = \sum_{k=0}^{n} \begin{pmatrix}n\\k\end{pmatrix} (2k)(4^{k})5$$

Ok so I am short a 2 on the right hand side now...

 

[pupeye11]

nevermind found my algebra mistake. thanks for the help.