(in Summation notation)
SUMMATION[from k=0 to k=floor(n/2)] (n-k)C(k) * 2(n-k-1)
refer to this image
after expanding it becomes
(n)C(0)*2(n-1) + (n-1)C(1)*2(n-2) + (n-2)C(2) * 2(n-3) +....
I want to calculate the value of function for a large value of n(up to
1010, and ans is to be found MODULUS to some prime number..like
Please simplify the series if possible or let me know the way how to
solve it efficiently.