#### Answer

The required solution is ${{\left( a+b \right)}^{n}}=\left( \begin{align}
& n \\
& 0 \\
\end{align} \right){{a}^{n}}+\underline{\left( \begin{align}
& n \\
& 1 \\
\end{align} \right)}{{a}^{n-1}}\cdot b+\underline{\left( \begin{align}
& n \\
& 2 \\
\end{align} \right)}{{a}^{n-2}}\cdot {{b}^{2}}+\underline{\left( \begin{align}
& n \\
& 3 \\
\end{align} \right)}{{a}^{n-3}}\cdot {{b}^{3}}+...+\underline{\left( \begin{align}
& n \\
& n \\
\end{align} \right)}\cdot {{b}^{n}}$

#### Work Step by Step

The sum of exponents on a and b in each term is n.
While writing the expansion of ${{\left( a+b \right)}^{n}}$, we note the following points:
The first term of the expansion of ${{\left( a+b \right)}^{n}}$ is ${{a}^{n}}$. And the exponents on a decrease by 1 in each successive term.
And the exponents on b in the expansion of ${{\left( a+b \right)}^{n}}$ increase by 1 in each successive term. In the first term, the exponent on b is 0 as ${{b}^{0}}=1$. And the last term is ${{b}^{n}}$.
And the sum of the exponents on the variables in any term in the expansion of ${{\left( a+b \right)}^{n}}$ is equal to n.
So, the total number of terms in the polynomial expansion is one more than the power of the binomial, i.e., n. And there are (n+1) terms in the expanded form of ${{\left( a+b \right)}^{n}}$.
Hence, for any positive integers n,