Question

5. sum_{i=0}^{n} (frac{1}{3})^i = A. frac{3}{2} - (frac{1}{3})^n B. frac{3}{2} [1 - (frac{1}{3})^n] C. frac{3}{2} (frac{1}{3})^n D. frac{2}{3} (frac{1}{3})^n E. frac{2}{3} (frac{1}{3})^{n-1}

          5. sum_{i=0}^{n} (frac{1}{3})^i = 

A. frac{3}{2} - (frac{1}{3})^n 

B. frac{3}{2} [1 - (frac{1}{3})^n] 

C. frac{3}{2} (frac{1}{3})^n 

D. frac{2}{3} (frac{1}{3})^n 

E. frac{2}{3} (frac{1}{3})^{n-1}

$5. sumi=0^n (frac13)^i = A. frac32 - (frac13)^n B. frac32 [1 - (frac13)^n] C. frac32 (frac13)^n D. frac23 (frac13)^n E. frac23 (frac13)^n-1$

Added by Connie A.

The IIT Foundation series of Mathematics Class

Trishna Knowledge Systems 5th Edition

Chapter 7

Instant Answer

Solved by Expert Sam Stansfield

05/10/2024

Step 1

The sum of an infinite geometric series is given by the formula $S = \frac{a}{1 - r}$, where $S$ is the sum of the series, $a$ is the first term, and $r$ is the common ratio. Substituting the given values into the formula, we get: \(S = Show more…

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