Question

Consider the infinite series sigma ^infinity _k = 1 1/7^k a. Find the first four terms of the sequence of partial sums. b. Use the results of part (a) to propose a formula for S_n. c. Propose a value of the series. a. Find the first four terms of the sequence of partial sums. S_1 =, S_2 =, S_3 =, S_4 = (Simplify your answers.) b. Use the results of part (a) to propose a formula for S_n. A. S_n = 7^n - 1/6 middot 7^n B. S_n = 7n/7^n C. S_n = n/7^n D. S_n = 6 middot 7^n/7^n - 1 c. Propose a value of the series. A reasonable conjecture for the value of the series is

Answer 1

Answer:

a. First four terms of the sequential partial sums

$S_1=\frac{1}{7}, S_2=\frac{7^2-1}{6*7^2}, S_3 =\frac{7^3-1}{6*7^3}, S_4 =\frac{7^4-1}{6*7^4}$

b. The formula for Sn is option A = $\frac{7^n-1}{6*7^n}$

c. Value of the series = $\frac{1}{6}$

Step-by-step explanation:

a. First four terms of the sequential partial sums

$\sum^{\infty}_{k=1}\\S_1=\frac{1}{7}\\ S_2=\frac{1}{7}+ \frac{1}{7^2}= \frac{8}{49} =\frac{7^2-1}{6*7^2}\\S_3=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3} = \frac{57}{343} =\frac{7^3-1}{6*7^3}\\S_4=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4} = \frac{400}{2401} =\frac{7^4-1}{6*7^4}$

b. Formula for Sn

The sum of n terms

$S_n=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+ \frac{1}{7^4}+.....+ \frac{1}{7^n}= \frac{7^n-1}{6*7^n}$

c. Value for the series

This can be given as the Sum of infinite terms;

$S_{\infty}=\frac{1}{7}+ \frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+....+ \frac{1}{7^n}+ ....= \lim_{n \to \infty} \frac{7^n-1}{6*7^n}= \frac{1}{6}$