Question

n^th term of the series
\(1+\frac{3}{7}+\frac{5}{7^2}+\frac{1}{7^2}+...\) is

• A. \(\frac{2n-1}{7^n}\)
• B. \(\frac{2n-1}{7^{n-1}}\)
• C. \(\frac{2n+1}{7^{n-1}}\)
• D. \(\frac{2n+1}{7^n}\)

Answer 1

The Correct Option is B

The correct answer is (V) :\(\frac{2n-1}{7^{n-1}}\) .

Answer 2

The correct answer is: (V) \(\frac{2n-1}{7^{n-1}}\).

The given series is:

\(1 + \frac{3}{7} + \frac{5}{7^2} + \frac{1}{7^2} + ...\)

We need to find the n^th term of the series. Observing the series, we see that the numerators follow the pattern: 1, 3, 5, ..., which are consecutive odd numbers. The denominators are powers of 7: \( 7^{0}, 7^1, 7^2, 7^3, \dots \). 
So, the general expression for the n^th term can be written as:

\(\frac{2n-1}{7^{n-1}}\).

This formula represents the n^th term of the series, where the numerator is the 2n - 1 (the odd numbers) and the denominator is the corresponding power of 7, starting from \(7^{0}\).