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I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.

The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have

$\sum_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$

when $N$ is odd.  (When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.

Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$,

$ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $

for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.

For fixed $k$ and $N \to \infty$, note that $$ {{n \choose k} + {n \choose k-1} + {n \choose k-2}+\dots \over {n \choose k}} = {1 + {k \over n-k+1} + {k(k-1) \over (n-k+1)(n-k+2)} + \cdots} $$ and we can bound the right side from above by the geometric series $$ {1 + {k \over n-k+1} + \left( {k \over n-k+1} \right)^2 + \cdots} $$ which equals ${n-(k-1) \over n - (2k-1)}$. Therefore we have $$ f(n,k) \le {n \choose k} {n-(k-1) \over n-(2k-1)}. $$

I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.

The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have

$\sum_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$

when $N$ is odd. 
(When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.

Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$.
Then you'll have, for real constant $\alpha$,

$ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $

for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.

For fixed $k$ and $N \to \infty$, note that $$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots \over {N \choose k}} = {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$ and we can bound the right side from above by the geometric series $$ {1 + {k \over N-k+1} + \left( {k \over N-k+1} \right)^2 + \cdots} $$ which equals ${N-(k-1) \over N - (2k-1)}$. Therefore we have $$ f(N,k) \le {N \choose k} {N-(k-1) \over N-(2k-1)}.$$

I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.

The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have

$\sum_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$

when $N$ is odd. (When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.

Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$,

$ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $

for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.

For fixed $k$ and $N \to \infty$, note that $$ {{n \choose k} + {n \choose k-1} + {n \choose k-2} \over {n \choose k}} = {1 + {k \over n-k+1} + {k(k-1) \over (n-k+1)(n-k+2)} + \cdots} $$ and we can bound the right side from above by the geometric series $$ {1 + {k \over n-k+1} + \left( {k \over n-k+1} \right)^2 + \cdots} $$ which equals ${n-(k-1) \over n - (2k-1)}$. Therefore we have $$ f(n,k) \le {n \choose k} {n-(k-1) \over n-(2k-1)}. $$

I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.

The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have

$\sum_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$

when $N$ is odd. (When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.

Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$,

$ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $

for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.

For fixed $k$ and $N \to \infty$, note that $$ {{n \choose k} + {n \choose k-1} + {n \choose k-2}+\dots \over {n \choose k}} = {1 + {k \over n-k+1} + {k(k-1) \over (n-k+1)(n-k+2)} + \cdots} $$ and we can bound the right side from above by the geometric series $$ {1 + {k \over n-k+1} + \left( {k \over n-k+1} \right)^2 + \cdots} $$ which equals ${n-(k-1) \over n - (2k-1)}$. Therefore we have $$ f(n,k) \le {n \choose k} {n-(k-1) \over n-(2k-1)}. $$

I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.

The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have

$\sum_{i=0}^{(N-1)/2} {N \choose k} = {2^N \over 2} = 2^{N-1}$

when $N$ is odd. (When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.

Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$,

$ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $

for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.

For fixed $k$ and $N \to \infty$, note that $$ {{n \choose k} + {n \choose k-1} + {n \choose k-2} \over {n \choose k}} = {1 + {k \over n-k+1} + {k(k-1) \over (n-k+1)(n-k+2)} + \cdots} $$ and we can bound the right side from above by the geometric series $$ {1 + {k \over n-k+1} + \left( {k \over n-k+1} \right)^2 + \cdots} $$ which equals ${n-(k-1) \over n - (2k-1)}$. Therefore we have $$ f(n,k) \le {n \choose k} {n-(k-1) \over n-(2k-1)}. $$

I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.

The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have

$\sum_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$

when $N$ is odd. (When $N$ is even something similar is true but you have to correct for whether you include the term ${N \choose N/2}$ or not.

Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$,

$ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $

for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables.

For fixed $k$ and $N \to \infty$, note that $$ {{n \choose k} + {n \choose k-1} + {n \choose k-2} \over {n \choose k}} = {1 + {k \over n-k+1} + {k(k-1) \over (n-k+1)(n-k+2)} + \cdots} $$ and we can bound the right side from above by the geometric series $$ {1 + {k \over n-k+1} + \left( {k \over n-k+1} \right)^2 + \cdots} $$ which equals ${n-(k-1) \over n - (2k-1)}$. Therefore we have $$ f(n,k) \le {n \choose k} {n-(k-1) \over n-(2k-1)}. $$