Question

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

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

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

Added by Connie A.

Mathematical Methods in the Physical Sciences

Mary L. Boas 2nd Edition

Chapter 1

Instant Answer

Solved by Expert Zhumagali Shomanov

05/10/2024

Step 1

The sum of an infinite geometric series can be found using the formula S = a / (1 - r), where a is the first term and r is the common ratio. In this case, the first term a is (1/3)^1 = 1/3 and the common ratio r is 1/3. Show more…

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