Problem 55
Question
Find the indicated term of each binomial expansion. \((q-3)^{9} ;\) second term
Step-by-Step Solution
Verified Answer
The second term of the binomial expansion \((q-3)^9\) is \(-27q^8\).
1Step 1: Determine the values of a, b, n, and k
In this binomial expansion, the values are:
a=q,
b=-3,
n=9,
k=1 (since we want to find the second term)
2Step 2: Calculate the binomial coefficient
Using the formula for the binomial coefficient \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) , let's calculate \(\binom{9}{1}\):
\[\binom{9}{1} = \frac{9!}{1!(9-1)!} = \frac{9!}{1!8!}\]
Since \(9! = 9 \times 8!\), we can simplify the expression:
\[\frac{9!}{1!8!} = \frac{9 \times 8!}{1! \times 8!} = 9\]
3Step 3: Substitute the values into the binomial theorem formula
Using the binomial theorem formula and the values we've determined, we have:
\[\text{2nd term} = \binom{9}{1} \times q^{9-1} \times (-3)^1 = 9 \times q^8 \times (-3)\]
4Step 4: Simplify to find the second term
Simplify the expression
\[\text{2nd term} = 9 \times q^8 \times (-3) = -27q^8\]
The second term of the binomial expansion \((q-3)^9\) is \(-27q^8\).
Other exercises in this chapter
Problem 54
Write each series using summation notation. 11+\frac{11}{2}+\frac{11}{3}+\frac{11}{4}+\frac{11}{5}+\frac{11}{6}
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Find \(S_{8}\) for each arithmetic sequence described below. $$a_{1}=2, d=3$$
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Use the formula for \(S_{n}\) to find the sum of the terms of each geometric sequence. $$-\frac{1}{4},-\frac{1}{2},-1,-2,-4,-8$$
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Write each series using summation notation. 3+6+9+12
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