To determine the coefficient of the last term in the binomial expansion of [tex]\((x+1)^9\)[/tex], we need to consider the binomial expansion formula. The binomial theorem states:
[tex]\[
(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
\][/tex]
In this specific problem, our binomial expression is [tex]\((x+1)^9\)[/tex]. Here, [tex]\(x = x\)[/tex], [tex]\(y = 1\)[/tex], and [tex]\(n = 9\)[/tex]. We need to find the coefficient of the last term. In the binomial expansion, the terms are written from [tex]\(k=0\)[/tex] to [tex]\(k=n\)[/tex]. The last term corresponds to [tex]\(k = 9\)[/tex].
Substituting [tex]\(n=9\)[/tex] and [tex]\(k=9\)[/tex] into the binomial coefficient formula gives us:
[tex]\[
\binom{9}{9} x^{9-9} 1^9
\][/tex]
Let's break this down:
- [tex]\(\binom{9}{9}\)[/tex] is the binomial coefficient, which represents the number of ways to choose 9 items out of 9.
- [tex]\(x^{9-9}\)[/tex] simplifies to [tex]\(x^0\)[/tex], which is [tex]\(1\)[/tex] for any value of [tex]\(x\)[/tex].
- [tex]\(1^9\)[/tex] is simply [tex]\(1\)[/tex], since any number to the power of 0 is 1.
The binomial coefficient [tex]\(\binom{9}{9}\)[/tex] simplifies as follows:
[tex]\[
\binom{9}{9} = \frac{9!}{9!(9-9)!} = \frac{9!}{9! \cdot 0!} = \frac{9!}{9! \cdot 1} = 1
\][/tex]
Therefore, the coefficient of the last term in the expansion of [tex]\((x+1)^9\)[/tex] is:
[tex]\[
1
\][/tex]
Thus, the correct answer is:
[tex]\[
1
\][/tex]
