Find the sum to infinity to the following series 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
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Correct Answer: Option D
Explanation:
3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
a = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9
3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
a = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9