Answer
$\begin{align}
& {{\left( a+b \right)}^{k+1}}={{a}^{k+1}}+\sum\limits_{r=1}^{k}{\left( \left( \begin{matrix}
k \\
r \\
\end{matrix} \right)+\left( \begin{matrix}
k \\
r-1 \\
\end{matrix} \right) \right){{a}^{k-r+1}}{{b}^{r}}}+{{b}^{k+1}} \\
& ={{a}^{k+1}}+\sum\limits_{r=1}^{k}{\left( \begin{matrix}
k+1 \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+{{b}^{k+1}} \\
& ={{a}^{k+1}}+\left( \begin{matrix}
k+1 \\
1 \\
\end{matrix} \right){{a}^{k-1+1}}{{b}^{1}}+\left( \begin{matrix}
k+1 \\
2 \\
\end{matrix} \right){{a}^{k-2+1}}{{b}^{2}}+...........+{{b}^{k+1}} \\
& =\sum\limits_{r=0}^{k+1}{\left( \begin{matrix}
k+1 \\
r \\
\end{matrix} \right){{a}^{k+1-r}}{{b}^{r}}}
\end{align}$
Work Step by Step
Let us consider the provided formula:
${{\left( a+b \right)}^{n}}=\left( \begin{matrix}
n \\
0 \\
\end{matrix} \right){{a}^{n}}+\left( \begin{matrix}
n \\
1 \\
\end{matrix} \right){{a}^{n-1}}b+\left( \begin{matrix}
n \\
2 \\
\end{matrix} \right){{a}^{n-2}}{{b}^{2}}+......+\left( \begin{matrix}
n \\
n-1 \\
\end{matrix} \right)a{{b}^{n-1}}+\left( \begin{matrix}
n \\
n \\
\end{matrix} \right){{b}^{n}}$
The above formula resembles the binomial expansion formula as shown below,
${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{a}^{n-r}}{{b}^{r}}}$
The mathematical induction uses a method of placing the value of n as 1, k and $\left( k+1 \right)$ successively. In each substitution with a particular value of n, the value in the left hand side is resolved; if it becomes equal to the value in the right hand side then the above mentioned formula is said to be true. And the steps can be summarized as,
$\begin{align}
& \left( a \right)n=1 \\
& \left( b \right)n=k \\
& \left( c \right)n=k+1
\end{align}$
And in each step the term in the LHS must be equal to the RHS for the formula to be true. If this is satisfied then the formula is said to be true.
Now, put the value of n as 1 in the LHS of the above expansion formula and resolve.
$\begin{align}
& {{\left( a+b \right)}^{n}}={{\left( a+b \right)}^{1}} \\
& =\left( a+b \right)
\end{align}$
Now, put the value of n as 1 in the RHS of the above expansion formula and resolve.
$\begin{align}
& \sum\limits_{r=0}^{1}{\left( \begin{matrix}
1 \\
r \\
\end{matrix} \right){{a}^{1-r}}{{b}^{r}}}=\left( \begin{matrix}
1 \\
0 \\
\end{matrix} \right){{a}^{1-0}}{{b}^{0}}+\left( \begin{matrix}
1 \\
1 \\
\end{matrix} \right){{a}^{1-1}}{{b}^{1}} \\
& =a+b
\end{align}$
And the above value shows that LHS is equal to RHS when the value of n is 1.
Now, put the value of n as k in the above expansion formula.
${{\left( a+b \right)}^{k}}=\sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r}}{{b}^{r}}}$
Now, put the value of n as k in the LHS of the above expansion formula and resolve.
$\begin{align}
& {{\left( a+b \right)}^{n}}={{\left( a+b \right)}^{k}} \\
& ={{\left( a+b \right)}^{k}}
\end{align}$
Now, put the value of n as k in the RHS of the above expansion formula and resolve.
$\sum\limits_{r=0}^{n}{\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{a}^{n-r}}{{b}^{r}}}=\sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r}}{{b}^{r}}}$
Which shows that LHS is equal to RHS when the value of n is k. So, the expansion formula is true when n is equal to k.
Now, put the value of n as $\left( k+1 \right)$ in the above expansion formula.
${{\left( a+b \right)}^{k+1}}=\sum\limits_{r=0}^{k+1}{\left( \begin{matrix}
n \\
r \\
\end{matrix} \right){{a}^{k+1-r}}{{b}^{r}}}$
Then, consider LHS of the above equation and resolve.
$\begin{align}
& {{\left( a+b \right)}^{k+1}}={{\left( a+b \right)}^{k}}\left( a+b \right) \\
& =\left( \sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r}}{{b}^{r}}} \right)\left( a+b \right) \\
& =a\left( \sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r}}{{b}^{r}}} \right)+b\left( \sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r}}{{b}^{r}}} \right)
\end{align}$
Further simplify,
$\begin{align}
& {{\left( a+b \right)}^{k+1}}=\sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+\sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r}}{{b}^{r+1}}} \\
& =\sum\limits_{r=0}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+\sum\limits_{r=1}^{k+1}{\left( \begin{matrix}
k+1 \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}} \\
& =\left( \begin{matrix}
k \\
0 \\
\end{matrix} \right){{a}^{k+1}}+\sum\limits_{r=1}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+\sum\limits_{r=1}^{k}{\left( \begin{matrix}
k \\
r-1 \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+\left( \begin{matrix}
k+1 \\
k+1 \\
\end{matrix} \right){{b}^{k+1}} \\
& ={{a}^{k+1}}+\sum\limits_{r=1}^{k}{\left( \begin{matrix}
k \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+\sum\limits_{r=1}^{k}{\left( \begin{matrix}
k \\
r-1 \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+{{b}^{k+1}}
\end{align}$
Then, simplifying
$\begin{align}
& {{\left( a+b \right)}^{k+1}}={{a}^{k+1}}+\sum\limits_{r=1}^{k}{\left( \left( \begin{matrix}
k \\
r \\
\end{matrix} \right)+\left( \begin{matrix}
k \\
r-1 \\
\end{matrix} \right) \right){{a}^{k-r+1}}{{b}^{r}}}+{{b}^{k+1}} \\
& ={{a}^{k+1}}+\sum\limits_{r=1}^{k}{\left( \begin{matrix}
k+1 \\
r \\
\end{matrix} \right){{a}^{k-r+1}}{{b}^{r}}}+{{b}^{k+1}} \\
& ={{a}^{k+1}}+\left( \begin{matrix}
k+1 \\
1 \\
\end{matrix} \right){{a}^{k-1+1}}{{b}^{1}}+\left( \begin{matrix}
k+1 \\
2 \\
\end{matrix} \right){{a}^{k-2+1}}{{b}^{2}}+...........+{{b}^{k+1}} \\
& =\sum\limits_{r=0}^{k+1}{\left( \begin{matrix}
k+1 \\
r \\
\end{matrix} \right){{a}^{k+1-r}}{{b}^{r}}}
\end{align}$
So, it is equal to the RHS.
The above equation shows that the LHS is equal to the RHS. This shows that the expansion formula is also true for $\left( k+1 \right)$.
Hence, the mathematical induction describes the binomial expansion formula to be true.
