Brackets indicated a combination

[StellaLuna]

I'm quite stuck with how to approahc this type of question.

Σ(k=100 to 201) Σ(j=100 to k) (201 over k+1)(j over 100)

Sorr for the set up, it is tricky to type. The brackets indicated a combination.

 

[Werg22]

First, figure out what the inner sum is in function of k. Then see if you know how to sum up what you get over all values of k.

 

[StellaLuna]

Thank you for responding, I still don't quite know how to even start what you suggested.

 

[Werg22]

By the way, are you sure the index of j starts at 100? Also, do you want to find the exact value of the sum or simply an approximation?

 

[StellaLuna]

I'm looking for an expression involving one or two binomial coefficients.
And yes j starts at 100

 

[Mandark]

For the inner sum \sum_{j=100}^k \binom{j}{100} you will want to make use of the http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity#Hockey-Stick_Identity . Once you evaluate the inner sum you'll have a sum over a product of binomial coefficients, if you expand the coefficients as factorials you'll be able to write the sum as a product of a binomial coefficient which doesn't depend on k, and a familiar binomial coefficient which does. The resulting sum will be easy to evaluate. :)

 

[StellaLuna]

\sum_{k=100}^{201} \binom{201}{k+1} \binom{k+1}{101}

Is what I got after using the hockey stick identity. I then carried out both combinations but was not sure how to rearrange them after?

 

[Mandark]

\sum_{k=100}^{201} \binom{201}{k+1} \binom{k+1}{101}

= \sum_{k=100}^{200}\frac{201!}{(200-k)!(k+1)!} \cdot\frac{(k+1)!}{(k-100)!101!}

= \frac{201!}{100!\cdot 101!} \sum_{k=0}^{100} \frac{100!}{(100-k)! k!}

then convert back to binomial coefficients