## Precalculus (6th Edition) Blitzer

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}}
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,