evaluate:\n$$\\sum_{n=1}^{\\infty} 8\\left(\\frac{3}{4}\\right)^{n-1}$$\n$$s = ?$$\nsum of an infinite…

evaluate:\n$$\\sum_{n=1}^{\\infty} 8\\left(\\frac{3}{4}\\right)^{n-1}$$\n$$s = ?$$\nsum of an infinite geometric series, $s = \\frac{a}{1-r}$

evaluate:\n$$\\sum_{n=1}^{\\infty} 8\\left(\\frac{3}{4}\\right)^{n-1}$$\n$$s = ?$$\nsum of an infinite geometric series, $s = \\frac{a}{1-r}$

Answer

Explanation:

Step1: Identify the first term $a$

Substitute $n = 1$ into the expression $8(\frac{3}{4})^{n-1}$ to find the first term. $$a = 8\left(\frac{3}{4}\right)^{1-1} = 8\left(\frac{3}{4}\right)^0 = 8(1) = 8$$

Step2: Identify the common ratio $r$

The common ratio is the base of the exponential term. $$r = \frac{3}{4}$$

Step3: Check for convergence

Since $|r| = |\frac{3}{4}| < 1$, the infinite geometric series converges. $$|r| < 1$$

Step4: Apply the sum formula

Use the formula $S = \frac{a}{1 - r}$ with $a = 8$ and $r = \frac{3}{4}$. $$S = \frac{8}{1 - \frac{3}{4}}$$

Step5: Simplify the denominator

Calculate the difference in the denominator. $$1 - \frac{3}{4} = \frac{1}{4}$$

Step6: Calculate the final sum

Divide the numerator by the simplified denominator. $$S = \frac{8}{\frac{1}{4}} = 8 \times 4 = 32$$

Answer:

32