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how to solve the geometric series
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\[\sum_{k=0}^{\infty} \left( \frac{ 1 }{ 3 } \right)^{k} (5^{3-k})\]
can someone show me how to solve the series the answer is \[\frac{ 2500 }{ 19 }\]
honestly, I don't get 2500/19 maybe there is someone who can help you
\[5^3,\frac{5^2}{3},\frac{5}{3^2},......\]Here \(\frac{1}{15}\) is common ratio so \[s_{\infty}=\frac{a}{(1-r)}\]
\[a=5^3\]\[s_{\infty }=\frac{5^3}{(1-\frac{1}{15})}\]
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As Jitesh said, you can find the sum of geometric series like that. Anyway, the answer wouldn't be 2500/19 after you calculate it. Is 2500/19 really the true answer?
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