evaluate:\n$$\\sum_{n=1}^{\\infty} 9\\left(\\frac{5}{6}\\right)^{n-1}$$\n$$s = ?$$\nsum of an infinite…
evaluate:\n$$\\sum_{n=1}^{\\infty} 9\\left(\\frac{5}{6}\\right)^{n-1}$$\n$$s = ?$$\nsum of an infinite geometric series, $s = \\frac{a}{1-r}$
Answer
Explanation:
Step1: Identify the first term
The first term $a$ is found by substituting $n=1$ into the expression $9(\frac{5}{6})^{n-1}$. $$a = 9\left(\frac{5}{6}\right)^{1-1} = 9(1) = 9$$
Step2: Identify the common ratio
The common ratio $r$ is the base of the exponential part of the geometric series. $$r = \frac{5}{6}$$
Step3: Apply the infinite sum formula
Substitute $a = 9$ and $r = \frac{5}{6}$ into the formula $S = \frac{a}{1-r}$. $$S = \frac{9}{1 - \frac{5}{6}}$$
Step4: Simplify the denominator
Calculate the difference in the denominator. $$1 - \frac{5}{6} = \frac{1}{6}$$
Step5: Calculate the final sum
Divide the numerator by the simplified denominator. $$S = \frac{9}{\frac{1}{6}} = 9 \times 6 = 54$$
Answer:
54