Weird Function | CodeChef

[ratnesht23]

CodeChef: Practical coding for everyone

how to solve this kind of problem. what kind of knowledge i need to solve this kind of question

[everule1]

refer to:

CODIGO 2k20, Weird Function Help

\sum_0^n\binom{n}{i}=2^n and (a+b)^n = \sum_0^n \binom{n}{i}a^ib^{n-i} so when m=1 \sum_0^n\sum_0^{n-i}\binom{n}{i}\binom{n-i}{j} =\sum_0^n \binom{n}{i} \times \sum_0^{n-i}\binom{n-i}{j} since \binom{n}{i} is independent of j. =\sum_0^n \binom{n}{i} \times 2^{n-i} =(1+2)^n = 3^n now when m=2 f(n,2)=\sum_i^n \binom{n}{i}f(n-i,1) =\sum_i^n \binom{n}{i}3^{n-i} =(3+1)^n = 4^n by recursion, f(n,m)=(m+2)^n \prod_{1}^{n} f(i,m) =\prod_{1}^{n} (m+2)^i =(m+2)^{\sum_0^n i} =(m+2)^\frac{n*…

[ratnesht23]

thank you so much really helped