Solve Sum of {30 \choose i} with Binomial Theorem

[mathgirl1]

Simplify (find the sum) of $${30 \choose 0} + \frac{1}{2}{30 \choose 1}+ \frac{1}{3}{30 \choose 2} + ... + \frac{1}{31}{30 \choose 30}$$.

Do this is two ways:
1. Write $$\frac{1}{i+1}{30 \choose i}$$ in a different way then add
2. Integrate the binomial thorem (don't forget the constant of integration)

I know the Bionomial Theorem is $$(x+y)^n = \sum_{k=0}^{n} {{n \choose k}x^ky^{n-k}}$$. So obviously I have y=1 and n=30 but I don't know how to convert $$\frac{1}{1+i}$$ into the form $$x^k$$or $$y^{n-k}$$ for all values of k. Can anyone help with this? I'm sure if I can figure out how to write it in the form I need then I can compute the sum using $$(x+y)^n$$ and then integrate this but need some help. Any help is much appreciated. Thank you!

 

[mathgirl1]

I know that $$x=(k+1)^{\frac{-1}{k}}$$ but I don't know how to use this to compute $$(x+y)^n$$ since x is in terms of k and not in terms of x. Help please! I am sure this should be simple but I am stuck

 

[MarkFL]

Let's look at the first part of the problem...and let's generalize by using $n$ instead of 30:

I think I would begin by writing the sum $S$ using sigma notation:

$$S=\sum_{k=0}^{n}\left(\frac{1}{k+1}{n \choose k}\right)$$

Now, if we use the definition:

$${n \choose r}\equiv\frac{n!}{r!(n-r)!)}$$

We may write:

$$S=\sum_{k=0}^{n}\left(\frac{1}{k+1}\cdot\frac{n!}{k!(n-k)!}\right)$$

And then:

$$S=\frac{1}{n+1}\sum_{k=0}^{n}\left(\frac{(n+1)!}{(k+1)!((n+1)-(k+1))!}\right)$$

And then:

$$S=\frac{1}{n+1}\sum_{k=1}^{n+1}\left(\frac{(n+1)!}{k!(((n+1)-k)!}\right)$$

And then:

$$S=\frac{1}{n+1}\left(\sum_{k=0}^{n+1}\left(\frac{(n+1)!}{k!((n+1)-k)!}\right)-1\right)$$

And then:

$$S=\frac{1}{n+1}\left(\sum_{k=0}^{n+1}\left({n+1 \choose k}\right)-1\right)$$

Can you continue by applying the binomial theorem?

 

[MarkFL]

To follow up, we have:

$$2^n=(1+1)^n=\sum_{k=0}^{n}\left({n \choose k}1^{n-k}1^{k}\right)=\sum_{k=0}^{n}\left({n \choose k}\right)$$

Thus, there results:

$$S=\frac{2^{n+1}-1}{n+1}$$

Now, if we begin with the binomial theorem:

$$(1+x)^n=\sum_{k=0}^{n}\left({n \choose k}x^{k}\right)$$

And then integrate both sides w.r.t $x$, we obtain:

$$\frac{(1+x)^{n+1}}{n+1}+C=\sum_{k=0}^{n}\left({n \choose k}\frac{x^{k+1}}{k+1}\right)$$

To solve for the constant of integration $C$, let's use $x=0$:

$$\frac{(1+0)^{n+1}}{n+1}+C=\sum_{k=0}^{n}\left({n \choose k}\frac{0^{k+1}}{k+1}\right)$$

And from this, we obtain:

$$C=-\frac{1}{n+1}$$

Hence:

$$\frac{(1+x)^{n+1}-1}{n+1}=\sum_{k=0}^{n}\left({n \choose k}\frac{x^{k+1}}{k+1}\right)$$

And then letting $x=1$, we have our result:

$$S=\frac{2^{n+1}-1}{n+1}$$