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

[mathgirl1]

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

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

I know the Bionomial Theorem is \(\displaystyle (x+y)^n = \sum_{k=0}^{n} {{n \choose k}x^ky^{n-k}}\). So obviously I have y=1 and n=30 but I don't know how to convert \(\displaystyle \frac{1}{1+i}\) into the form \(\displaystyle x^k\)or \(\displaystyle y^{n-k}\) 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 \(\displaystyle (x+y)^n\) and then integrate this but need some help. Any help is much appreciated. Thank you!

 

[mathgirl1]

I know that \(\displaystyle x=(k+1)^{\frac{-1}{k}}\) but I don't know how to use this to compute \(\displaystyle (x+y)^n\) 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:

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

Now, if we use the definition:

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

We may write:

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

And then:

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

And then:

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

And then:

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

And then:

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

Can you continue by applying the binomial theorem?

 

[MarkFL]

To follow up, we have:

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

Thus, there results:

\(\displaystyle S=\frac{2^{n+1}-1}{n+1}\)

Now, if we begin with the binomial theorem:

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

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

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

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

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

And from this, we obtain:

\(\displaystyle C=-\frac{1}{n+1}\)

Hence:

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

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

\(\displaystyle S=\frac{2^{n+1}-1}{n+1}\)