## Precalculus (6th Edition) Blitzer

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}}
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,