MHB Prove Probability: Step-by-Step Guide

[Niamh1]

View attachment 6593

Prove the following

I literally have no idea where to start or what to do.

 

[Jameson]

Hi there. :)

The binomial theorem states that $$(x+y)^n = \sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}$$. What happens if you let $x=1$ and $y=1$?

 

[Niamh1]

Hi there. :)

The binomial theorem states that $$(x+y)^n = \sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}$$. What happens if you let $x=1$ and $y=1$?

Doesn't really help me to be honest.

 

[Euge]

Hi Niamh,

Let $S = \sum_{r = 1}^n r\binom{n}{r}$. Then

$$S = 0\binom{n}{0} + 1\binom{n}{1} + 2\binom{n}{2} + \cdots + (n-1)\binom{n}{n-1} + n\binom{n}{n}$$
$$S = n\binom{n}{n} + (n-1)\binom{n}{n-1} + \cdots + 1\binom{n}{1} + 0\binom{n}{0}$$

Using the fact that $\binom{n}{r} = \binom{n}{n-r}$ for $0 \le r \le n$, add the two equations column-wise, then use the binomial theorem, to obtain $2S = n2^n$. Dividing by $2$ yields the result.

 

[johng1]

Here's another easy proof using calculus:

$$(x+1)^n=\sum_{r=0}^n{{n}\choose{r}}x^r$$

Take derivatives:

$$n(x+1)^{n-1}=\sum_{r=1}^nr{{n}\choose{r}}x^{r-1}$$

Eavaluate at x=1:

$$n2^{n-1}=\sum_{r=1}^nr{{n}\choose{r}}=\sum_{r=0}^nr{{n}\choose{r}}$$

 

[Jameson]

Doesn't really help me to be honest.

Hi Niamh,

My post and the two below show the main connection you need, that is how to connect the summation sign to something resembling $2^n$. At first glance this is a quite strange equation and the key is the binomial theorem as I mentioned. If you let $x=1$ and $y=1$ you get:

$$(x+y)^n =(1+1)^n=2^n= \sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}=\sum_{k=0}^{n}\binom{n}{k}1^k1^{n-k}=\sum_{k=0}^{n}\binom{n}{k}$$

Using this trick we have now connected $2^n$ to something with this summation. That was the idea behind my post. :)

What have you tried?