A summation with binomial coefficients Evaluate



There seemed to be some problem with stackexchange's math rendering but Ian corrected whatever error was there in the expression.Thanks

$$5050 \frac {\left( \sum _{r=0}^{100} \frac {{100\choose r}}{50r+1}\cdot (-1)^r\right) - 1}{\left( \sum _{r=0}^{101}\frac{{101\choose r}}{50r+1} \cdot (-1)^r\right) - 1}$$

The original definite integral that led to this was 

$$5050\frac{\int_0^1(1-x^{50})^{100} dx}{\int_0^1(1-x^{50})^{101} dx} $$

I used binomail to arrive at the above expression.

I already know a technique by definite integration but I want one by sequence and series.


   The answer is 5051
Use of sequence and series is not suggested; and a possible way is outlined, which is very long and useless, this is due to the fact of bounded multiplication of r with 50 in denominator, if the r would have been free, it would be easy to use this method to caluculate:

Brute Force

$$\frac{100\choose r}{50r+1}=1-50.\frac{r.{100\choose r}}{50r+1}$$
Use: $$r{100\choose r}=100.{99\choose{r-1}}$$
Then
$$\frac{{99\choose{r-1}}}{50r+1}=\left[1-50.\frac{r.{99\choose {r-1}}}{50r+1}\right]$$
Again use:
$$r{99\choose {r-1}}=99.{98\choose{r-2}}\text{ after }\frac{r.{99\choose {r-1}}}{50r+1}=\frac{(r-1).{99\choose {r-1}}}{50r+1}+\frac{{99\choose {r-1}}}{50r+1}$$
and so on...



The way you are mentioning- definite integration of $\beta$-function (though you may not be realising it is the $\beta$-function)- is the best method in my opinion.



You should use these(preferred but not necessary) while summing binomial coefficients:
$$r{n\choose r}=n{{n-1}\choose{r-1}};\frac1r{n\choose r}=\frac1{n+1}{{n+1}\choose{r+1}}$$
Secondly, use partial fractions or partial rationals.
This expression can be simplified using Beta and Gamma functions. 
The sum in the numerator is
\begin{align}
\sum^{100}_{r=0}\binom{100}{r}\frac{(-1)^r}{50r+1}
&amp;=\sum^{100}_{r=0}\binom{100}{r}(-1)^r\int^1_0x^{50r}dx\\
&amp;=\int^1_0(1-x^{50})^{100}dx\\
&amp;=\frac{1}{50}\int^1_0u^{-49/50}(1-u)^{100}du\\
&amp;=\frac{1}{50}B(1/50,101)\\
&amp;=\frac{\Gamma(1/50) \ \Gamma(101)}{50 \ \Gamma(5051/50)}
\end{align}
Similarly, the sum in the denominator is
\begin{align}
\int^1_0(1-x^{50})^{101}dx
&amp;=\frac{\Gamma(1/50) \ \Gamma(102)}{50 \ \Gamma(5101/50)}
\end{align}
Assuming you meant
$$5050\frac{\sum^{100}_{r=0}\binom{100}{r}\frac{(-1)^r}{50r+1}}{\sum^{101}_{r=0}\binom{101}{r}\frac{(-1)^r}{50r+1}}$$
instead, the above expression evaluates to
$$\frac{5050 \ \color\red{\Gamma(1/50)} \ \Gamma(101)}{\color\red{50} \ \Gamma(5051/50)}\frac{\color\red{50} \ \Gamma(5101/50)}{\color\red{\Gamma(1/50)} \ \Gamma(102)}=\frac{5050\cdot5051}{50\cdot 101}=5051$$