Answer
The required solution is ${{\left( x+2 \right)}^{5}}=\left( \begin{align}
& 5 \\
& 0 \\
\end{align} \right){{x}^{5}}+\underline{\left( \begin{align}
& 5 \\
& 1 \\
\end{align} \right)}{{x}^{4}}\cdot 2+\underline{\left( \begin{align}
& 5 \\
& 2 \\
\end{align} \right)}{{x}^{3}}\cdot {{2}^{2}}+\underline{\left( \begin{align}
& 5 \\
& 3 \\
\end{align} \right)}{{x}^{2}}\cdot {{2}^{3}}+\underline{\left( \begin{align}
& 5 \\
& 4 \\
\end{align} \right)}x\cdot {{2}^{4}}+\underline{\left( \begin{align}
& 5 \\
& 5 \\
\end{align} \right)}\cdot {{2}^{5}}$
Work Step by Step
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 ${{a}^{n}}$. The exponents 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 ${{b}^{0}}=1$. 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.
And the total number of terms in the polynomial expansion is one more than the power of the binomial, i.e., n. There are (n+1) terms in the expanded form of ${{\left( a+b \right)}^{n}}$.
Hence, for any positive integers n,
