A Fraction And A Binomial

We know that $(1+x)^n=1+\binom n1 x+\binom n2 x^2+\dots +x^n$ Integrating both the sides w.r.t $x$ we get $\frac{(1+x)^{n+1}}{n+1}+\mathbb C=x+\frac 12\binom n1 x^2+\frac 13\binom n2 x^3+\dots +\frac 1{n+1} x^{n+1}$ Substituting $x=0$ in the above expression we get $\mathbb C=-\dfrac 1{n+1}$ .

$\frac{(1+x)^{n+1}}{n+1}-\frac 1{n+1}=x+\frac 12\binom n1 x^2+\frac 13\binom n2 x^3+\dots +\frac 1{n+1}x^{n+1}$

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Now substituting $x=1$ and $x=-1$ and adding the equations gives us

$\begin{aligned} \dfrac{2^{n+1}-2}{n+1}&=2\left(\frac 12\binom n1+\frac 13\binom n2+\cdots +\frac 1{n+1}\right)\\ \dfrac{2^{n}-1}{n+1}&=\frac 12\binom n1+\frac 13\binom n2+\cdots +\frac 1{n+1}\\ \end{aligned}$

Now substituting $n=30$ we get $\mathbb E = \frac{2^{30}-1}{31}$

Thus $a+b+c = 62$ .

First we write the series in sigma notation:

$E=\sum\limits_{n=1}^{15}{\dfrac{1}{2n} \binom{30}{2n-1}}$

The approach here is to try and simplify the summand as far as possible:

$=\sum\limits_{n=1}^{15}{\dfrac{1}{2n}\dfrac{30!}{(31-2n)!(2n-1)!}}$

$=\dfrac{1}{31}\sum\limits_{n=1}^{15}{\dfrac{31\cdot 30!}{(2n)!(31-2n)!}}$

$=\dfrac{1}{31}\sum\limits_{n=1}^{15}{\binom{31}{2n}}$

From the Binomial Theorem:

$(x+y)^{n}=\sum\limits_{k=0}^{n}{\binom{n}{k} x^{k}y^{n-k}}$

We take look at when $x=y=1$ and $x=1$ and $y=-1$ , and $n=31$ :

$2^{31}=\sum\limits_{k=0}^{31}{\binom{31}{k}{(1)^k}{(1)^{31-k}}}$ _ _ _ _ _ (1)

$0=\sum\limits_{k=0}^{31}{\binom{31}{k}{(1)^k}{(-1)^{31-k}}}$ _ _ _ _ _ (2)

Then we subtract the two equations 1 and 2, to get:

$2^{31}=\sum\limits_{k=0}^{31}{\binom{31}{k}{(1-(-1)^{31-k})}}$

$2^{31}=\sum\limits_{k=0}^{31}{\binom{31}{k}{(1+(-1)^{-k})}}$

Now we collect the non-zero terms and re-write this in sigma notation to get:

$2^{31}=2+2\sum\limits_{k=1}^{15}{\binom{31}{2k}}$

$\therefore \sum\limits_{k=1}^{15}{\binom{31}{2k}}=2^{30}-1$

We substitute this back in $E$ to get:

$E=\dfrac{1}{31}(2^{30}-1)=\dfrac{2^{30}-1}{31}$

$\therefore a+b+c=30+1+31=62$