Question :
Find the value of the following expression :
$$ \frac{\sum_{i=0}^{2024}\sum_{r=0}^{2024}(-1)^r{2024 \choose r}(2024-r)^i}
{\sum_{r=0}^{2025}(-1)^r\binom{2025}{r}(2025-r)^{2025}} $$
I am not able to think of a way to manipulate and solve this algebraically , so I decided to test this on smaller summation boundaries to find a pattern in the answers For ex :
$$ \frac{\sum_{i=0}^{2}\sum_{r=0}^{2}(-1)^r{2 \choose r}(2-r)^i}
{\sum_{r=0}^{3}(-1)^r\binom{3}{r}(3-r)^{3}} $$
On substituting 2024 with 2 and 2025 with 3 and solving the resultant expression (summing each case) I got the answer 2/6 which reduced to 1/3 . Here I observed that 3 is the upper boundary for the summation in the denominator , so I checked for 2 more cases which turned out to be quite tedious but :
- 2024 –> 3 and 2025 –> 4 Answer – 6/24 –> 1/4
- 2024 –> 4 and 2025 –> 5 Answer – 24/120 –> 1/5
A pattern is formed where the reciprocal of the upper boundary of the summation in denominator is the answer . There is also the pattern of factorials where 2 is factorial of 2 , 6 is of 3 , 24 is of 4 and 120 is of 5 . So I can guess the answer to my original question to be 1/2025 following the pattern , which also turns out to be the correct answer
But this is definitely not how the question was intended to be solved , so I need a way to solve this with just algebraic manipulations in mind.
There could be a way to generalize this as :
$$ \frac{\sum_{i=0}^{k}\sum_{r=0}^{k}(-1)^r{k \choose r}(k-r)^i}
{\sum_{r=0}^{k+1}(-1)^r\binom{k+1}{r}(k+1-r)^{k+1}} = \frac{k!}{(k+1)!}=\frac{1}{k+1} $$
Best Answer
Let $$N_k:=\frac{\sum_{i=0}^{k}\sum_{r=0}^{k}(-1)^r{k \choose r}(k-r)^i} {\sum_{r=0}^{k+1}(-1)^r\binom{k+1}{r}(k+1-r)^{k+1}} =\frac{\sum_{i=0}^ka_{i,k}}{a_{k+1,k+1}}$$ where $$a_{k,i}:=\sum_{r=0}^k(-1)^r\binom kr(k-r)^i.$$ In the proposed duplicate and posts linked to it, it is proved that $$a_{k,k}=k!.$$ In this other post and posts linked to it, it is proved that $$\forall i<k\quad a_{k,i}=0.$$ Therefore, $$N_k=\frac{k!+0}{(k+1)!}=\frac1{k+1}.$$