Prove by mathematical induction:
$\sum_{i=0}^{n}{n \choose i} = 2^n ; n \ge 0$
Step 1: n = 0
${0 \choose 0}=2^0$
Step 2:
for n = k
$\sum_{i=0}^{k}{k \choose i} = 2^k$
assumption: for n = k+1
$\sum_{i=0}^{k+1}{k+1 \choose i} = 2^{k+1}$
Step 3:
$\sum_{i=0}^{k+1}{k+1 \choose i} = \sum_{i=0}^{k}{k \choose i} + {k+1 \choose k+1} = 2^k+1$
Assumption:
$2^{k+1}$
Result of mathematical induction:
$2^k+1$
Is that correct or is somewhere the mistake ?
Your base case is essentially correct, although I would prefer to see the justification that both sides are equal to $1$. In step 2, your assumption should be $$\sum_{i = 0}^{k} \binom{k}{i} = 2^k$$ since you must show that $P(k + 1)$ holds whenever $P(k)$ holds. In step 3, the statement $$\sum_{i = 0}^{k + 1} \binom{k + 1}{i} = \sum_{i = 0}^{k} \binom{k}{i} + \binom{k + 1}{k + 1}$$ is false. Furthermore, if $k$ is a positive integer, $2^{k + 1} = 2 \cdot 2^k = 2^k + 2^k > 2^k + 1$.
An induction proof is provided below:
Proof. Let $P(n)$ be the statement that $$\sum_{k = 0}^{n} \binom{n}{k} = 2^n$$
Let $n = 0$. Then $$\sum_{k = 0}^{0} \binom{0}{k} = \binom{0}{0} = 1 = 2^0$$ Hence, $P(0)$ holds.
Since $P(0)$ holds, we may assume there exists a nonnegative integer $m$ such that $P(m)$ holds. Then $$\sum_{k = 0}^{m} \binom{n}{k} = 2^m$$ Let $n = m + 1$. We must show that $P(m) \Rightarrow P(m + 1)$.
\begin{align*} \sum_{k = 0}^{m + 1} \binom{m + 1}{k} & = \binom{m + 1}{0} + \sum_{k = 1}^{m} \binom{m + 1}{k} + \binom{m + 1}{m + 1}\\ & = 1 + \sum_{k = 1}^{m} \binom{m + 1}{k} + 1\\ & = 1 + \sum_{k = 1}^{m} \left[\binom{m}{k} + \binom{m}{k - 1}\right] + 1 & \text{by Pascal's Identity}\\ & = 1 + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{k = 1}^{m} \binom{m}{k - 1} + 1\\ & = 1 + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{j = 0}^{m - 1} \binom{m}{j} + 1 & \text{where $j = k - 1$}\\ & = \binom{m}{0} + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{j = 0}^{m - 1} \binom{m}{j} + \binom{m}{m}\\ & = \sum_{k = 0}^{m} \binom{m}{k} + \sum_{j = 0}^{m} \binom{m}{j}\\ & = 2^m + 2^m & \text{induction hypothesis}\\ & = 2 \cdot 2^{m}\\ & = 2^{m + 1} \end{align*} Since $P(0)$ holds and $P(m) \Rightarrow P(m + 1)$ for each nonnegative integer $m$, $P(n)$ holds for all nonnegative integers.$\blacksquare$
Note. The binomial coefficient $$\binom{n}{k}$$ represents the number of subsets of size $k$ that can be selected from a set of size $n$. Thus, the summation $$\sum_{k = 0}^{n} \binom{n}{k}$$ counts the number of subsets of a set with $n$ elements. There are $2^n$ such subsets since a subset is determined by the decision of whether or not to include each of the $n$ elements.