Bounding a sum of binomial coefficients I would like to bound the sum 
$$ \phi(n,k) = {n \choose k} + 2 {n-1 \choose k} + 3 {n-2 \choose k} + \cdots + (n-k+1) {k \choose k} $$
I'm trying to get a handle on how big this sum is. Specifically, how big is it as a function of $k$ when $n=k^{1.5}$? What about $n=k^2$? I need to obtain these kinds of estimates, and I'm wondering whether any one can suggest the best methods for this. 
 A: $$S~=~\sum_{j=k}^n{j\choose k}(n-j+1)~=~\sum_{i=0}^{n-k}{k+i\choose k}\Big[(n-k+1)-i\Big]~=~(n-k+1)~S_0-S_1,$$
where $~S_a~=~\displaystyle\sum_{i=0}^{n-k}{k+i\choose k}~i^a.\quad$ At the same time, $~\displaystyle{k+i\choose k}={k+i\choose i}=(-1)^i~{-k-1\choose i},$ 
and $~\displaystyle{a\choose i}~i~=~{a-1\choose i-1}~a.\quad$ Then, letting $a=-(k+1),~$ and using the well-known identity 
$\displaystyle\sum_{i=0}^m(-1)^i{p\choose i}~=~(-1)^m{p-1\choose m},\quad$ we $($finally$)$ arrive at $~S=\displaystyle{n+2\choose k+2}.\quad$ QED.
A: $$\begin{align}
\phi(n,k)&=\sum_{j=k}^n\binom jk\binom {n+1-j}1\\
&=\sum_{j=k}^n \binom j{j-k}\binom {n+1-j}{n-j}\\
&=\sum_{j=k}^n (-1)^{j-k}\binom {-k-1}{j-k}(-1)^{n-j}\binom{-2}{n-j}&&\text{(Upper Negation)}\\
&=(-1)^{n-k}\sum_{j=k}^n \binom {-k-1}{j-k}\binom{-2}{n-j}\\
&=(-1)^{n-k}\binom {-k-3}{n-k}&&\text{(Vandermonde)}\\
&=(-1)^{2(n-k)}\binom {n+2}{n-2}&&\text{(Upper Negation)}\\
&=\binom {n+2}{k+2}\quad\blacksquare
\end{align}$$
A: The above sum may be simplified to $\frac{(n+1)(n+2)\binom{n}{k}}{(k+1)(k+2)}$
Sorry, I did not have a chance to provide the explanation earlier.
Note that we are required to evaluate the following sum:
$$\sum_{t=0}^{n-k} (t+1) \binom{n-t}{k} = \sum_{t=0}^{n} \binom{t+1}{1} \binom{n-t}{k}$$
In the above sum, all the terms corresponding to $t =n-k$ to $t=n$ are zero.
Now consider the following identity:
$$\sum_{k = 0}^l \binom{l-k}{m} \binom{q+k}{n} = \binom{l+q+1}{m+n+1}$$ where $l, m, n, q \geq 0$
Let us use the above identity: plug in with $(k,l,m,n,q) \leftarrow (t,n,k,1,1)$ 
Hence the answer is $$\binom{n+2}{k+2} = \frac{(n+1)(n+2)\binom{n}{k}}{(k+1)(k+2)}$$
