Proving $\sum_{i=0}^{\rho} \binom{n}{i} (q-1)^i \leq (n q)^p$: Is It Possible?

[evinda]

Hello! (Wave)

I want to show that $\sum_{i=0}^{\rho} \binom{n}{i} (q-1)^i \leq (n q)^p$.

I thought to use the fact that $\sum_{i=0}^{\rho} \binom{n}{i} (q-1)^i \leq \sum_{i=0}^{\rho} \binom{\rho}{i} (nq-1)^i$.

I tried to prove this but for $n=1$ it doesn't hold. Does it hold for greater $n$ ?

Or do we have to use something else?

 

[Ackbach]

Hello! (Wave)

I want to show that $\sum_{i=0}^{\rho} \binom{n}{i} (q-1)^i \leq (n q)^p$.

I have three questions about this inequality:

1. Did you mean $\binom{n}{i}$, or did you mean $\binom{\rho}{i}$, or did you mean $\binom{p}{i}?$ If you did mean $\binom{n}{i}$, are there any constraints on the size of $n$, particularly relative to $\rho ?$

2. On the RHS, did you mean $(nq)^p$ or $(nq)^{\rho}?$

3. Are there any constraints on the size of $q$?