What is the coefficient of the last term in the binomial expansion of $(x+1)^9$ ?
To find the coefficient of the last term in the binomial expansion of \((x+1)^9\), we recall the Binomial Theorem. The Binomial Theorem states that for any non-negative integer \(n\), the expansion of \((a+b)^n\) is given by:
\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
In our specific case, \(a = x\), \(b = 1\), and \(n = 9\). Thus, the expansion is:
\[(x+1)^9 = \sum_{k=0}^{9} \binom{9}{k} x^{9-k} 1^k = \sum_{k=0}^{9} \binom{9}{k} x^{9-k}\]
The expansion of \((x+1)^9\) consists of terms for \(k = 0, 1, 2, \dots, 9\). The "last" term in this expansion corresponds to the highest value of \(k\), which is \(k = 9\). Substituting \(k = 9\) into the general form of the terms \(\binom{n}{k} x^{n-k} b^k\), we get:
\[\binom{9}{9} x^{9-9} 1^9 = \binom{9}{9} x^0 \cdot 1 = \binom{9}{9} \cdot 1 \cdot 1\]
The coefficient of the last term is given by the binomial coefficient \(\binom{9}{9}\). By definition, \(\binom{n}{n} = \frac{n!}{n!(n-n)!} = \frac{n!}{n!0!} = 1\). Therefore:
\[\binom{9}{9} = 1\]
The last term is \(1 \cdot x^0 = 1\), and its coefficient is 1. Comparing this to the provided options, the correct answer is b.
Binomial expansions are essential in various fields, such as probability and statistics, where they are used to model the number of successes in a sequence of independent experiments. For instance, when calculating the probability of getting exactly 9 heads in 9 flips of a coin, the binomial expansion helps determine the weight of the specific event. Understanding these coefficients allows for efficient calculations in fields ranging from engineering to finance, where predicting outcomes based on multiple scenarios is required.