+46 The largest term in the binomial expansion of \((1+\dfrac{1}{2})^{31}\) is of the form \(\dfrac{a}{b}\), where \(a\) and \(b\) are relatively prime positive integers. What is the value of \(b\)?
Please help! Thank you so much! All I know is to use the binomial theorem!
+704 The largest term in the binomial expansion of (a+b)n occurs when k = n/2 (for even n) or k = (n - 1)/2 (for odd n) in the binomial coefficient \binom{n}{k} = \frac{n!}{k!(n-k)!}.
In this case, n = 31 (odd). Therefore, the largest term occurs when k = (31 - 1)/2 = 15.
The term we want is \binom{31}{15} = \frac{31!}{15!16!}. We can re-write this as:
\binom{31}{15} = \frac{31 \times 30 \times 29 \times \cdots \times 16}{16 \times 15 \times 14 \times \cdots \times 1}
Notice that most of the terms in the numerator and denominator cancel out, leaving us with:
\binom{31}{15} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 15450
The denominator, b, is thus 3 \times 2 \times 1 = \boxed{6}.
