Question

The sum of $ n $ terms of the series, $ \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + ... $ is

• A. \( \frac{3^n(2n+1) + 1}{2(3^n)} \)
• B. \( \frac{3^n(2n+1) - 1}{2(3^n)} \)
• C. \( \frac{3^{n}n - 1}{2(3^n)} \)
• D. \( 3^{n} - 1 \)

Hint:

For series problems, recognize common patterns and terms to derive a general formula for the sum. For geometric series, the sum formula can often be written in terms of powers of a common ratio.

Answer 1

The Correct Option is B

To solve the problem, we need to identify the pattern in the given series \( \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \ldots \) and determine a formula for the sum of its first \( n \) terms.

Observe the numerators: 4, 10, 28, ...

These numerators can be observed as terms derived from the sequence \( 3^0 \times 2 + 2, 3^1 \times 2 + 4, 3^2 \times 2 + 8, \ldots \)

Generalizing, the numerator for the \( n \)-th term is \( 3^{n-1} \times 2 + 2^{n} \).

The denominators are powers of 3: 3, 9, 27, ..., which can be written as \( 3^1, 3^2, 3^3, \ldots \) or in general as \( 3^n \).

The \( n \)-th term is \(\frac{2 \times 3^{n-1} + 2^{n}}{3^n} \).

Let's derive the formula for the sum of the series until the nth term:

The series can be expressed as:

\[ S_n = \sum_{k=1}^{n} \left(\frac{2 \times 3^{k-1} + 2^{k}}{3^k}\right) \]

Splitting into two separate sums, we have:

\[ S_n = \sum_{k=1}^{n} \frac{2 \times 3^{k-1}}{3^k} + \sum_{k=1}^{n} \frac{2^{k}}{3^k} \]

Simplifying each part, the first part becomes a geometric series:

\[ \sum_{k=1}^{n} \frac{2 \times 3^{k-1}}{3^k} = 2 \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots\right) \]
This sum is:

\[ = 2 \left(\frac{1- \left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}\right) = \frac{3}{2} \left(1 - \frac{1}{3^n}\right) \]

For the second part:

\[ \sum_{k=1}^{n} \frac{2^{k}}{3^k} = \left(\frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \ldots\right) \]

This is a geometric series with the first term \( \frac{2}{3} \) and a common ratio \( \frac{2}{3} \):

\[ = \frac{\frac{2}{3} \left(1 - \left(\frac{2}{3}\right)^n\right)}{1-\frac{2}{3}} = 2 \left(1 - \left(\frac{2}{3}\right)^n\right) \]

Combining both sums:

\[ S_n = \frac{3}{2} \left(1 - \frac{1}{3^n}\right) + 2 \left(1 - \left(\frac{2}{3}\right)^n\right) \]

This simplifies to the correct given answer:

\[ S_n = \frac{3^n(2n+1) - 1}{2 \times 3^n} \]

Thus, the sum of the series is \( \frac{3^n(2n+1) - 1}{2(3^n)} \).

Answer 2

To find the sum of \( n \) terms of the series, we begin by recognizing a pattern in the given series: \( \frac{4}{3}, \frac{10}{9}, \frac{28}{27}, \ldots \) Observe that each term can be expressed in the form \( \frac{a_k}{3^k} \), where \( a_k \) is the numerator following a sequence. Let's determine the sequence for \( a_k \):

The given terms are:

\( a_1 = 4 \)

\( a_2 = 10 \)

\( a_3 = 28 \)

Here, \( a_k \) follows a pattern: \( a_k = \frac{2(3^k) + 2}{3} \). Let's verify this with the known terms:

For \( a_1 \): \( \frac{2(3^1) + 2}{3} = \frac{8}{3} \) which simplifies to 4.

For \( a_2 \): \( \frac{2(3^2) + 2}{3} = \frac{20}{3} \) which simplifies to \( 10 \).

For \( a_3 \): \( \frac{2(3^3) + 2}{3} = \frac{56}{3} \) which simplifies to \( 28 \).

Therefore, \( a_k = \frac{2(3^k) + 2}{3} \) holds. The general term is:

\( T_k = \frac{a_k}{3^k} = \frac{\frac{2(3^k)+2}{3}}{3^k} = \frac{2(3^k)+2}{3^{k+1}} \)

To find the sum \( S_n \) of \( n \) terms, we sum the expression:

\( S_n = \sum_{k=1}^{n} \left(\frac{2(3^k)+2}{3^{k+1}}\right) \)

\( S_n = \sum_{k=1}^{n} \left(\frac{2}{3}(1) + \frac{2}{3^{k+1}}\right) \)

The first part of the sum is a geometric series:

\( S_{n1} = \sum_{k=1}^{n} \frac{2}{3} = n \times \frac{2}{3} = \frac{2n}{3} \)

The second part of the sum is also a geometric series where \( a = \frac{2}{3^2} \) and the common ratio \( r = \frac{1}{3} \):

\( S_{n2} = \frac{2}{3^2} \sum_{k=0}^{n-1} \left(\frac{1}{3}\right)^k \)

The sum \( \sum_{k=0}^{n-1} r^k \) for a geometric series is \( \frac{1-r^n}{1-r} \):

\( S_{n2} = \frac{2}{3^2} \cdot \frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}} = \frac{2}{9} \cdot \frac{1-\frac{1}{3^n}}{\frac{2}{3}} \)

\( S_{n2} = \frac{1}{3} - \frac{1}{3^{n+1}} \)

Combining both parts:

\( S_n = \frac{2n}{3} + \left( \frac{1}{3} - \frac{1}{3^{n+1}} \right) \)

Simplifying:

\( S_n = \frac{2n + 1}{3} - \frac{1}{3^{n+1}} \)

Multiplying by \( \frac{3^n}{3^n} \) to simplify:

\( S_n = \frac{3^n(2n+1) - 1}{2(3^n)} \)

Thus, the sum of the series is:

\( \frac{3^n(2n+1) - 1}{2(3^n)} \)