- The binomial coefficients for \((a+b)^8\) are determined by \(\binom{8}{k}\) for \(k = 0, 1, \dots, 8\).
- Using the binomial formula, we calculate the sequence: \(1, 8, 28, 56, 70, 56, 28, 8, 1\).
- Comparing this sequence with the given options, we identify option d as the correct one.
- The final answer is \(\boxed{d}\).
Explanation
- Analysis of the Problem and Binomial Theorem
The binomial expansion of \((a+b)^n\) is given by the formula \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) are the binomial coefficients. For \(n=8\), the expression is \((a+b)^8 = \sum_{k=0}^{8} \binom{8}{k} a^{8-k} b^k\), which results in 9 terms. We need to identify the list of 9 coefficients: \(\binom{8}{0}, \binom{8}{1}, \binom{8}{2}, \binom{8}{3}, \binom{8}{4}, \binom{8}{5}, \binom{8}{6}, \binom{8}{7}, \binom{8}{8}\).
- Calculating the Coefficients
To find the coefficients, we calculate each value using the formula \(\binom{8}{k} = \frac{8!}{k!(8-k)!}\):
- \(\binom{8}{0} = \frac{8!}{0!8!} = 1\)
- \(\binom{8}{1} = \frac{8!}{1!7!} = 8\)
- \(\binom{8}{2} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28\)
- \(\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
- \(\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\)
Using the symmetry property \(\binom{n}{k} = \binom{n}{n-k}\), the remaining coefficients are:
- \(\binom{8}{5} = \binom{8}{3} = 56\)
- \(\binom{8}{6} = \binom{8}{2} = 28\)
- \(\binom{8}{7} = \binom{8}{1} = 8\)
- \(\binom{8}{8} = \binom{8}{0} = 1\)
The complete sequence is \(1, 8, 28, 56, 70, 56, 28, 8, 1\).
- Comparison and Conclusion
Comparing the calculated sequence \(1, 8, 28, 56, 70, 56, 28, 8, 1\) with the provided options:
A\(1, 8, 28, 56, 70, 70, 56, 28, 8, 1\)
B\(1, 7, 21, 35, 21, 7, 1\)
C\(1, 7, 21, 35, 35, 21, 7, 1\)
D\(1, 8, 28, 56, 70, 56, 28, 8, 1\)
Option (d) matches the sequence of coefficients exactly. Thus, the coefficients in the binomial expansion of \((a+b)^8\) are \(\boxed{d}\).
Examples
The binomial coefficients are essential in probability and combinatorics. For instance, if you are flipping a coin 8 times, the coefficients of \((a+b)^8\) help determine the number of ways to get exactly \(k\) heads. By setting \(a=b=1\), we see that the sum of the coefficients \(\binom{8}{0} + \binom{8}{1} + \dots + \binom{8}{8} = 2^8 = 256\), which is the total number of possible outcomes when flipping 8 coins. This helps in understanding the distribution of outcomes in various real-world scenarios.