Answer
False statement
Work Step by Step
${{\left( x+y \right)}^{^{17}}}$.
The binomial theorem, for any binomial $\left( a+b \right)$ and any natural number $n$,
${{\left( a+b \right)}^{n}}={{c}_{0}}{{a}^{n}}{{b}^{0}}+{{c}_{1}}{{a}^{n-1}}{{b}^{1}}+{{c}_{2}}{{a}^{n-2}}{{b}^{2}}+\cdots +{{c}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{{c}_{n}}{{a}^{0}}{{b}^{n}}$
Where the constants ${{c}_{0}},{{c}_{1}},{{c}_{2}},\ldots ,{{c}_{n}}$ are known as the binomial coefficient.
And the number of terms in the expression after combining like terms in the expansion is $\left( n+1 \right)$ terms.
The given expression is ${{\left( x+y \right)}^{^{17}}}$.
Compare ${{\left( x+y \right)}^{^{17}}}$ to ${{\left( a+b \right)}^{n}}$ in which $n=17$
Thus, the number of terms N is,
$\begin{align}
& N=\left( n+1 \right) \\
& =\left( 17+1 \right) \\
& =18
\end{align}$
Therefore, the expansion of ${{\left( x+y \right)}^{^{17}}}$has 18 terms not 19.
Thus, the given statement is false.
