# Brackets indicated a combination

1. Feb 7, 2010

### 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.

2. Feb 7, 2010

### Werg22

Re: sum

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.

3. Feb 7, 2010

### StellaLuna

Re: sum

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

4. Feb 7, 2010

### Werg22

Re: sum

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?

5. Feb 7, 2010

### StellaLuna

Re: sum

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

6. Feb 7, 2010

### Mandark

Re: sum

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 [Broken]. 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. :)

Last edited by a moderator: May 4, 2017
7. Feb 8, 2010

### StellaLuna

Re: sum

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

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?

8. Feb 8, 2010

### Mandark

Re: sum

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

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

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

then convert back to binomial coefficients