Geometric sum with binomial coefficient I'm looking for any kind of formula or insight concerning
$$
f(k,N)=\sum_{n=0}^N\binom{n}{k}\lambda^n
$$
There are loads of binomial identities, so maybe I was just missing something, but I couldn't find anything yet.

Addendum:
Further assumptions on $\lambda$, i.e. $\lambda=\exp(2\pi i\alpha)$ yield to the following problem: Gaussian like sum with binomial coefficients
 A: Allow me to change your notation to
$$
f(x,m,n) = \sum\limits_{k = 0}^n {\binom{k}{m}x^{\,k} } 
$$
Then a first approach would be through a double o.g.f. as
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {f(x,m,n)y^n }
  = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k\, \le \,n}
 {\left( \matrix{  k \cr   m \cr}  \right)x^{\,k} y^n } }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {\sum\limits_{\,k\, \le \,n}
 {\left( \matrix{  k \cr   m \cr}  \right)\left( {xy} \right)^{\,k} y^{n - k} } }
  = \sum\limits_{0\, \le \,k} {\sum\limits_{\,0\, \le \,j}
 {\left( \matrix{  k \cr   m \cr}  \right)\left( {xy} \right)^{\,k} y^j } }  =   \cr 
  &  = {1 \over {1 - y}}\sum\limits_{0\, \le \,k}
 {\left( \matrix{  k \cr   k - m \cr}  \right)\left( {xy} \right)^{\,k} }
  = {{\left( {xy} \right)^{\,m} } \over {1 - y}}\sum\limits_{0\, \le \,k - m}
 {\left( { - 1} \right)^{\,k - m} \left( \matrix{   - m - 1 \cr   k - m \cr}  \right)
\left( {xy} \right)^{\,k - m} }  =   \cr 
  &  = {{\left( {xy} \right)^{\,m} } \over {\left( {1 - y} \right)\left( {1 - xy} \right)^{\,m + 1} }} \cr} 
$$
An other way is first to express the sum as an infinite sum with the index starting from $0$, as
$$
\eqalign{
  & f(x,m,n) = \sum\limits_{k = 0}^n {\left( \matrix{  k \cr   m \cr}  \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {\left( \matrix{  n - k \cr   n - k \cr}  \right)
\left( \matrix{  k \cr   k - m \cr}  \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,j}
 {\left( \matrix{  j \cr   j \cr}  \right)\left( \matrix{  n - j \cr  n - m - j \cr}  \right)x^{\,n - j} }  =   \cr 
  &  = x^{\,n} \sum\limits_{0\, \le \,j} {{{\left( {n - j} \right)^{\,\underline {\,n - m - j\,} } }
 \over {\left( {n - m - j} \right)!}}\left( {1/x} \right)^{\,j} }  \cr} 
$$
where $x^{\,\underline {\,k\,} }$
represents the Falling Factorial
Then we can convert it into a Hypergeometric function by putting
$$
t_{\,j}  = {{\left( {n - j} \right)^{\,\underline {\,n - m - j\,} } } \over {\left( {n - m - j} \right)!}}
$$
and determine that
$$
t_{\,0}  = {{\left( n \right)^{\,\underline {\,n - m\,} } } \over {\left( {n - m} \right)!}}
 = \left( \matrix{  n \cr   m \cr}  \right)
$$
as well that the ratio is
$$
\eqalign{
  & {{t_{\,j + 1} } \over {t_{\,j} }}
 = {{\left( {n - 1 - j} \right)^{\,\underline {\,n - 1 - m - j\,} } }
 \over {\left( {n - 1 - m - j} \right)!}}{{\left( {n - m - j} \right)!} 
\over {\left( {n - j} \right)^{\,\underline {\,n - m - j\,} } }} =   \cr 
  &  = {{\left( {n - 1 - j} \right)^{\,\underline {\,n - 1 - m - j\,} } }
 \over {\left( {n - j} \right)^{\,\underline {\,1\,} } 
\left( {n - 1 - j} \right)^{\,\underline {\,n - 1 - m - j\,} } }}{{\left( {n - m - j} \right)} \over 1} =   \cr 
  &  = {{n - m - j} \over {n - j}} = {{j - n + m} \over {j - n}} \cr} 
$$
Therefore
$$
f(x,m,n) = x^{\,n} \left( \matrix{  n \cr   m \cr}  \right){}
_2F_{\,1} \left( {\left. {\matrix{   { - n + m,\;1}  \cr    { - n}  \cr  } \;}
 \right|\;{1 \over x}} \right)
$$
Still another way to express $f(x,m,n)$ is
$$
\eqalign{
  & f(x,m,n) = \sum\limits_{k = 0}^n {\left( \matrix{  k \cr  m \cr}  \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_{k = 0}^n {{{k^{\,\underline {\,m\,} } } \over {m!}}x^{\,k} }
  = {{x^{\,m} } \over {m!}}\sum\limits_{k = 0}^n {k^{\,\underline {\,m\,} } x^{\,k - m} }  =   \cr 
  &  = {{x^{\,m} } \over {m!}}\sum\limits_{k = 0}^n {{{d^{\,m} } \over {dx^{\,m} }}x^{\,k} }
  = {{x^{\,m} } \over {m!}}{{d^{\,m} } \over {dx^{\,m} }}
\left( {{{1 - x^{\,n + 1} } \over {1 - x}}} \right) \cr} 
$$
A: To get a sum of about $k$ terms (opposed to $N$ originally), use $$\sum_{n=0}^N\binom{n}{k}\lambda^n=\frac{\lambda^k}{k!}\frac{d^k}{d\lambda^k}\sum_{n=0}^N\lambda^n=\frac{\lambda^k}{k!}\frac{d^k}{d\lambda^k}\frac{1-\lambda^{N+1}}{1-\lambda}$$ and apply Leibniz rule. This can also be viewed as a closed-form answer for a fixed $k$.
