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sum_(k=1)^(oo)4((1)/(3))^(k-1)

k=14(13)k1 \sum_{k=1}^{\infty} 4\left(\frac{1}{3}\right)^{k-1}

Full solution

Q. k=14(13)k1 \sum_{k=1}^{\infty} 4\left(\frac{1}{3}\right)^{k-1}
  1. Recognize series type: We recognize that the series is a geometric series with the first term a=4 a = 4 (since when k=1 k = 1 , the term is 41 4 \cdot 1 ) and the common ratio r=13 r = \frac{1}{3} . The sum of an infinite geometric series can be found using the formula S=a1r S = \frac{a}{1 - r} , where |r| < 1 for the series to converge.
  2. Check convergence: We check the common ratio to ensure that the series converges. Since |r| = |\frac{1}{3}| = \frac{1}{3} < 1 , the series converges.
  3. Apply sum formula: We apply the formula for the sum of an infinite geometric series: S=a1r S = \frac{a}{1 - r} . Substituting a=4 a = 4 and r=13 r = \frac{1}{3} , we get S=4113 S = \frac{4}{1 - \frac{1}{3}} .
  4. Simplify denominator: We simplify the denominator: 113=3313=23 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} .
  5. Calculate sum: We calculate the sum: S=423 S = \frac{4}{\frac{2}{3}} . To divide by a fraction, we multiply by its reciprocal: S=432 S = 4 \cdot \frac{3}{2} .
  6. Perform multiplication: We perform the multiplication: S=432=23=6 S = 4 \cdot \frac{3}{2} = 2 \cdot 3 = 6 .

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