I am trying to prove the following bound using induction (where N>D):

$\sum_{i=0}^{D}\binom{N}{i} \leq N^{D} + 1$

I would appreciate any help on how to prove this. Thanks.

Currently I have looked at low values of D (base cases) but I don't understand how to formulate an induction step without using any prior information on the final result.

D=0: $\sum_{i=0}^{0}\binom{N}{i} \leq N^{0} + 1 \Leftrightarrow 1 \leq 2$

D=1: $\sum_{i=0}^{1}\binom{N}{i} \leq N^{1} + 1 \Leftrightarrow N + 1 \leq N + 1$

D=2: $\sum_{i=0}^{2}\binom{N}{i} \leq N^{2} + 1 \Leftrightarrow \frac{N^{2}}{2} + \frac{N}{2} + 1 \leq \frac{N^{2}}{2} + \frac{N^{2}}{2} + 1 \leq N^{2} + 1$

Maybe a first step in solving this is:

$\sum_{i=0}^{D}\binom{N}{i} = \sum_{i=0}^{D-1}\binom{N}{i} + \binom{N}{D}$ ??

  • $\begingroup$ I edited the post to show what I have tried. Do you have any suggestions on how to formulate an induction step? $\endgroup$ – ahmad1374 Feb 14 '17 at 4:00
  • $\begingroup$ You are allowed to use the prior information for the D case to prove D+1. $\endgroup$ – Q the Platypus Feb 14 '17 at 5:20

Possible Solution Route?

I'll change notation because I like using $a,n$ more $$1+ a^n \ge \sum_{k=0}^n {a \choose k} \tag{1}$$ Assuming you have your base case, we assume $(1)$. We now have that $$1+(a+1)^n$$ $$=1+\sum_{k=0}^n {n \choose k}a^k$$ $$=1+a^n+\sum_{k=0}^{n-1} {n \choose k}a^k$$ $$\ge\sum_{k=0}^n {a \choose k}+\sum_{k=0}^{n-1} {n \choose k}a^k$$

We now note that $$\sum_{k=0}^{n} {a+1 \choose k}= \sum_{k=0}^{n} \left[{a \choose k}+{a \choose k-1}\right]$$ This mean all we have to prove is that $$\sum_{k=0}^{n-1} {n \choose k}a^k \ge \sum_{k=0}^{n}{a \choose k-1}$$

  • $\begingroup$ Sorry I don't understand how proving the last statement can prove the original problem. Can you clarify? $\endgroup$ – ahmad1374 Feb 14 '17 at 12:44

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