show $\sum_{k=0}^n {k \choose i} = {n+1 \choose i+1}$ show for n $\geq i \geq 1 : \sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$
i show this with induction:
for n=i=1: ${1+1} \choose {1+1}$ = $2 \choose 2$ = 1 = $0 \choose 1$ + $1 \choose 1$ = $\sum_{k=0}^1 {k \choose 1}$
now let $\sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$
for n+1: ${n+1+1} \choose {i+1}$ = $n+1 \choose i$ + $n+1 \choose i$ = $ n+1 \choose i$ + $\sum_{k=0}^n {k \choose i} = \sum_{k=0}^{n+1} {k \choose i}  $ 
Is this the right way ? 
 A: Induction on $n$ will work, but your argument got into trouble (perhaps only typo trouble) with indices. We want to show that if the result is true for $n=w$, then it is true for $n=w+1$.
So our induction assumption is that 
$$\sum_{k=0}^w\binom{k}{i}=\binom{w+1}{i+1},$$
and we want to show that
$$\sum_{k=0}^{w+1}\binom{k}{i}=\binom{w+2}{i+1},$$
Consider the sum
$$\sum_{k=0}^{w+1}\binom{k}{i}.\tag{1}$$
This is equal to 
$$\sum_{k=0}^{w}\binom{k}{i}+\binom{w+1}{i}.\tag{2}$$
By the induction assumption, we have
$$\sum_{k=0}^w \binom{k}{i}=\binom{w+1}{i+1}.$$
Thus (2) is equal to
$$\binom{w+1}{i+1}+\binom{w+1}{i}.\tag{3}$$
By the Pascal Identity, (3) is equal to
$$\binom{w+2}{i+1},$$
which is what we wanted to show.
A: We can also use the recurrence from Pascal's Triangle and telescoping series:
$$
\begin{align}
\sum_{k=0}^n\binom{k}{i}
&=\sum_{k=0}^n\left[\binom{k+1}{i+1}-\binom{k}{i+1}\right]\\
&=\sum_{k=1}^{n+1}\binom{k}{i+1}-\sum_{k=0}^n\binom{k}{i+1}\\
&=\binom{n+1}{i+1}-\binom{0}{i+1}\\
&=\binom{n+1}{i+1}
\end{align}
$$
