Given a polynomial $ f(x) = \sum_{n=0}^{200}a_{n}x^{n}$ satisfied with equation $f(x)+f(x-1) = (x+1)^{200}$ is true for all $x\in \mathbb{R}$
Question : What is the value of $1+ \frac{2 \sum_{n=51}^{100}\binom{2n-1}{100}a_{2n-1}}{\binom{200}{100}}$
i've started with gathering information about $a_{n}$ and noticed that
$a_{n} + \sum_{k=n}^{200}(-1)^{k}a_{k} = (-1)^{n}\binom{200}{200-n} \ ; \ n=0,1,2,3,...,200$
it's seem like we should find recurrence relation for $a_{n}$ and a bit of combinatoric sum work.
If it's has any easier solution , please give me some advice.
Thanks for your help.
Essentially, the task is to compute $\frac12(f^{(100)}(1)-f^{(100)}(-1))$, that is, the odd part of the 100th derivative evaluated at $x=1$. Another observation is that the arguments in this desired expression have difference $2$, while the defining equation is about the sum of values with argument difference $1$. This can be extended to argument difference $2$. $$ f(x+1)-f(x-1)=[f(x+1)+f(x)]-[f(x)+f(x-1)]=(x+2)^{200}-(x+1)^{200}. $$ The 100th derivative is then \begin{align} (f^{(100)}(x+1)-f^{(100)}(x-1) &= \sum_{k=100}^{199} k(k-1)...(k-99)a_k[(x+1)^{k-100}-(x-1)^{k-100}] \\ &=200⋅199⋅198...101[(x+2)^{100}-(x+1)^{100}] \end{align} or $$ \sum_{k=100}^{199}\binom{k}{100}a_k[(x+1)^{k-100}-(x-1)^{k-100}] =\binom{200}{100}[(x+2)^{100}-(x+1)^{100}] $$
The evaluation at $x=0$ gives $$ f^{(100)}(1)-f^{(100)}(-1)=2\sum_{k=51}^{100}\binom{2k-1}{100}a_{2k-1} =\binom{200}{100}[2^{100}-1] $$ and from here it is easy to find the value of the expression in the task.