Here, the given series is
\(\sum_{n=2}^\infty\frac{5^n}{12^n}\)
Let us expand the given series.
\(\sum_{n=2}^\infty\frac{5^n}{12^n}=\frac{5^2}{12^2}+\frac{5^3}{12^3}+\frac{5^4}{12^4}+...\) (up to \(\infty)\)
\(=(\frac{5}{12})^2+(\frac{5}{12})^3+(\frac{5}{12})^4+...\)
We clearly see that the given series is an infinite geometric series with first term \((\frac{5}{12})^2\) and common ratio \((\frac{5}{12})\).
Again, We know that the sum of an infinite geometric series with first term a(say) and common ratio \(r(|r|\leq1)\) is \(\frac{a}{1-r}\)
So, \(\sum_{n=2}^\infty\frac{5^n}{12^n}=\frac{\frac{5^2}{12^2}}{1-\frac{5}{12}}\)
\(=\frac{\frac{25}{144}}{\frac{7}{12}}\)
\(=\frac{25}{144}\times\frac{12}{7}\)
\(=\frac{25}{84}\)
Which is the required answer.
\(\sum_{n=2}^\infty\frac{5^n}{12^n}\)
Let us expand the given series.
\(\sum_{n=2}^\infty\frac{5^n}{12^n}=\frac{5^2}{12^2}+\frac{5^3}{12^3}+\frac{5^4}{12^4}+...\) (up to \(\infty)\)
\(=(\frac{5}{12})^2+(\frac{5}{12})^3+(\frac{5}{12})^4+...\)
We clearly see that the given series is an infinite geometric series with first term \((\frac{5}{12})^2\) and common ratio \((\frac{5}{12})\).
Again, We know that the sum of an infinite geometric series with first term a(say) and common ratio \(r(|r|\leq1)\) is \(\frac{a}{1-r}\)
So, \(\sum_{n=2}^\infty\frac{5^n}{12^n}=\frac{\frac{5^2}{12^2}}{1-\frac{5}{12}}\)
\(=\frac{\frac{25}{144}}{\frac{7}{12}}\)
\(=\frac{25}{144}\times\frac{12}{7}\)
\(=\frac{25}{84}\)
Which is the required answer.
