Understanding the Series and the Sum Formula

The problem provides a series and a formula for its sum. The series is given by:

\( S_n = 1 \cdot 3^1 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n \)

This is an arithmetic-geometric series where the arithmetic component is \(1, 2, 3, \dots, n\) and the geometric component is \(3^1, 3^2, 3^3, \dots, 3^n\). The formula provided for the sum is:

\( S_n = \frac{{\left( {2n - 1} \right){3^a} + b}}{4} \)

We are asked to find the values of \(a\) and \(b\). Based on the options provided, \(a\) and \(b\) are either constants or expressions involving \(n\).

Method 1: Verifying with Small Values of n

A straightforward approach to find \(a\) and \(b\) is to calculate the sum of the series for small values of \(n\) and compare it with the result from the given formula, substituting the options for \(a\) and \(b\).

Calculate \(S_n\) for small n:

• For \(n=1\): \(S_1 = 1 \cdot 3^1 = 3\)
• For \(n=2\): \(S_2 = 1 \cdot 3^1 + 2 \cdot 3^2 = 3 + 2 \cdot 9 = 3 + 18 = 21\)

Test each option for \(a\) and \(b\):

Let the given formula with specific \(a\) and \(b\) be \(F_n(a, b) = \frac{(2n - 1)3^a + b}{4}\). We check if \(F_n(a, b) = S_n\) for \(n=1\) and \(n=2\).

Option 4 provides values for \(a\) and \(b\) that match the calculated sum for both \(n=1\) and \(n=2\). This strongly suggests that \(a = n + 1\) and \(b = 3\).

Method 2: Deriving the Sum Formula

The sum of an arithmetic-geometric series \( S_n = \sum_{k=1}^n [a + (k-1)d]r^k \) can be derived using standard techniques, such as multiplication by the common ratio \(r\) and subtracting the series. For the series \( S_n = \sum_{k=1}^n k \cdot 3^k \), we have the arithmetic part \(k\) (with initial term 1 and common difference 1) and the geometric part \(3^k\) (with common ratio 3).

Let \( S_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n \). (Equation 1)

Multiply Equation 1 by the common ratio \(3\):

\( 3S_n = 1 \cdot 3^2 + 2 \cdot 3^3 + 3 \cdot 3^4 + \dots + (n-1) \cdot 3^n + n \cdot 3^{n+1} \) (Equation 2)

Subtract Equation 1 from Equation 2:

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

This subtraction method should be done term by term, aligning terms with the same power of 3:

\( 3S_n = \quad \quad 1 \cdot 3^2 + 2 \cdot 3^3 + \dots + (n-1) \cdot 3^n + n \cdot 3^{n+1} \)

\( S_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n \)

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

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

The terms in the parenthesis form a geometric series with first term \(a = 3^2 = 9\), common ratio \(r = 3\), and number of terms \(n-1\). The sum of this geometric series is \( \frac{a(r^{N} - 1)}{r - 1} = \frac{9(3^{n-1} - 1)}{3 - 1} = \frac{9(3^{n-1} - 1)}{2} \).

Substitute this back into the equation for \(2S_n\):

\( 2S_n = -3 + \frac{9(3^{n-1} - 1)}{2} + n \cdot 3^{n+1} \)

\( 4S_n = -6 + 9(3^{n-1} - 1) + 2n \cdot 3^{n+1} \)

\( 4S_n = -6 + 9 \cdot 3^{n-1} - 9 + 2n \cdot 3^{n+1} \)

\( 4S_n = -15 + 3^2 \cdot 3^{n-1} + 2n \cdot 3^{n+1} \)

\( 4S_n = -15 + 3^{n+1} + 2n \cdot 3^{n+1} \)

\( 4S_n = (1 + 2n)3^{n+1} - 15 \)

\( S_n = \frac{(2n + 1)3^{n+1} - 15}{4} \)

Wait, this does not match the form \(\frac{(2n - 1)3^a + b}{4}\). Let's recheck the subtraction step of the series or consider the differentiation of the geometric series method.

Alternative Derivation using Differentiation

Consider the geometric series sum: \( G_n(x) = \sum_{k=0}^n x^k = 1 + x + x^2 + \dots + x^n = \frac{x^{n+1} - 1}{x - 1} \) for \(x \ne 1\).

Differentiate with respect to \(x\):

\( \frac{d}{dx} G_n(x) = \sum_{k=1}^n kx^{k-1} = 1 + 2x + 3x^2 + \dots + nx^{n-1} \)

\( \frac{d}{dx} \left(\frac{x^{n+1} - 1}{x - 1}\right) = \frac{(n+1)x^n(x-1) - (x^{n+1} - 1)(1)}{(x-1)^2} = \frac{(n+1)x^{n+1} - (n+1)x^n - x^{n+1} + 1}{(x-1)^2} = \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \)

So, \( \sum_{k=1}^n kx^{k-1} = \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \).

The series we need to sum is \( \sum_{k=1}^n k \cdot 3^k = 3 \sum_{k=1}^n k \cdot 3^{k-1} \). Substitute \(x=3\) into the derived formula:

\( \sum_{k=1}^n k \cdot 3^{k-1} = \frac{n3^{n+1} - (n+1)3^n + 1}{(3-1)^2} = \frac{n3^{n+1} - (n+1)3^n + 1}{4} \)

Now, multiply by 3 to get \(S_n\):

\( S_n = 3 \cdot \frac{n3^{n+1} - (n+1)3^n + 1}{4} \)

\( S_n = \frac{3n3^{n+1} - 3(n+1)3^n + 3}{4} \)

Rewrite \(3n3^{n+1}\) as \(3n \cdot 3 \cdot 3^n = 9n \cdot 3^n\). Rewrite \(3(n+1)3^n\) as \((3n+3)3^n\).

\( S_n = \frac{9n \cdot 3^n - (3n+3)3^n + 3}{4} \)

\( S_n = \frac{(9n - (3n+3))3^n + 3}{4} \)

\( S_n = \frac{(9n - 3n - 3)3^n + 3}{4} \)

\( S_n = \frac{(6n - 3)3^n + 3}{4} \)

This still does not match the required form \(\frac{(2n - 1)3^a + b}{4}\). Let's check the geometric series sum formula application. The sum of \( \sum_{k=1}^n kx^k = x \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \).

Substitute \(x=3\):

\( S_n = 3 \frac{n3^{n+1} - (n+1)3^n + 1}{(3-1)^2} = 3 \frac{n3^{n+1} - (n+1)3^n + 1}{4} \)

\( S_n = \frac{3n3^{n+1} - 3(n+1)3^n + 3}{4} \)

\( S_n = \frac{n3^{n+2} - (3n+3)3^n + 3}{4} \)

Let's try to match the form \(\frac{(2n - 1)3^{n+1} + 3}{4}\) from Option 4 with the derived sum. \(\frac{(2n - 1)3^{n+1} + 3}{4} = \frac{(2n - 1) \cdot 3 \cdot 3^n + 3}{4} = \frac{(6n - 3)3^n + 3}{4}\). Comparing this with \( \frac{n3^{n+2} - (3n+3)3^n + 3}{4} \): \( (6n - 3)3^n + 3 \) vs \( n3^{n+2} - (3n+3)3^n + 3 \) \( (6n - 3)3^n \) vs \( n \cdot 3^2 \cdot 3^n - (3n+3)3^n \) \( (6n - 3)3^n \) vs \( 9n \cdot 3^n - (3n+3)3^n \) \( (6n - 3)3^n \) vs \( (9n - 3n - 3)3^n \) \( (6n - 3)3^n \) vs \( (6n - 3)3^n \). They match!

The sum of the series is indeed \( S_n = \frac{(6n - 3)3^n + 3}{4} \). We need to express this in the form \( \frac{(2n - 1)3^a + b}{4} \). We can factor out 3 from \((6n - 3)\): \( 6n - 3 = 3(2n - 1) \) So, \( S_n = \frac{3(2n - 1)3^n + 3}{4} \). Using the rule \(x^m \cdot x^p = x^{m+p}\), we have \(3 \cdot 3^n = 3^{1} \cdot 3^n = 3^{n+1}\).

Thus, \( S_n = \frac{(2n - 1)3^{n+1} + 3}{4} \).

Comparing this with the given form \( \frac{(2n - 1)3^a + b}{4} \), we can identify the values of \(a\) and \(b\):

• The power of 3 is \(n+1\), so \(a = n + 1\).
• The constant term added is \(3\), so \(b = 3\).

Therefore, \(a = n + 1\) and \(b = 3\).

Conclusion on Finding a and b

Both the method of testing small values and the formal derivation of the sum formula lead to the same result for \(a\) and \(b\). The values are \(a = n + 1\) and \(b = 3\).

Revision Table: Sum of Series Concepts

The series in the problem is of the form \( \sum_{k=1}^n k \cdot 3^k \), which is an arithmetic-geometric series.

Additional Information on Series Summation

Summing arithmetic-geometric series can be done using a technique similar to summing geometric series. If \(S_n = a_1 g_1 + a_2 g_2 + \dots + a_n g_n\), where \(a_k\) are terms of an arithmetic progression with common difference \(D\) and \(g_k\) are terms of a geometric progression with common ratio \(R\), then consider \(S_n\) and \(RS_n\). Subtracting \(RS_n\) from \(S_n\) often results in a geometric series, which is easier to sum.

For the series \( S_n = \sum_{k=1}^n k \cdot x^k \), the general sum is \( \frac{x(1-(n+1)x^n+nx^{n+1})}{(1-x)^2} \). Let's substitute \(x=3\) into this general formula:

\( S_n = \frac{3(1-(n+1)3^n+n3^{n+1})}{(1-3)^2} = \frac{3(1-(n+1)3^n+n \cdot 3 \cdot 3^n)}{(-2)^2} \)

\( S_n = \frac{3(1-(n+1)3^n+3n \cdot 3^n)}{4} = \frac{3(1+(-n-1+3n)3^n)}{4} = \frac{3(1+(2n-1)3^n)}{4} \)

\( S_n = \frac{3 + 3(2n-1)3^n}{4} = \frac{3 + (2n-1)3^{n+1}}{4} \)

This matches the form \( \frac{(2n - 1)3^{n+1} + 3}{4} \), confirming \(a=n+1\) and \(b=3\).