Key Concepts and Formulas

• Arithmetico-Geometric Progression (AGP): A series where each term is the product of a term from an Arithmetic Progression (AP) and a term from a Geometric Progression (GP). The general form is $S = a + (a+d)r + (a+2d)r^2 + \dots + (a+(n-1)d)r^{n-1}$.
• Sum of an AGP: The standard method involves multiplying the series by the common ratio ($r$) of the GP part and subtracting it from the original series. This results in a geometric series that can be summed using the GP sum formula.
• Sum of a Geometric Progression (GP): For a GP with first term $A$, common ratio $R$, and $N$ terms, the sum is $S_N = \frac{A(R^N - 1)}{R-1}$ (when $R \neq 1$).

Step-by-Step Solution

Step 1: Identify the Series and its Components The given series is $S = 1 + 2 \cdot 3 + 3 \cdot 3^2 + \dots + 10 \cdot 3^9$. We can rewrite this as: $S = 1 \cdot 3^0 + 2 \cdot 3^1 + 3 \cdot 3^2 + \dots + 10 \cdot 3^9 \quad \dots (1)$ This is an AGP with:

• Arithmetic Progression (AP): $1, 2, 3, \dots, 10$. Here, the first term $a=1$, common difference $d=1$, and the number of terms $n=10$.
• Geometric Progression (GP): $3^0, 3^1, 3^2, \dots, 3^9$. Here, the first term of the GP part is $1$ (or $3^0$), and the common ratio $r=3$. The number of terms in the GP part matches the number of terms in the AP, so $n=10$.

Step 2: Multiply the Series by the Common Ratio of the GP The common ratio of the geometric part is $r=3$. Multiply equation (1) by 3: $3S = 1 \cdot 3^1 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + 9 \cdot 3^9 + 10 \cdot 3^{10} \quad \dots (2)$ We align the terms in the next step for subtraction.

Step 3: Subtract the Original Series from the Multiplied Series Subtract equation (1) from equation (2). This is a standard technique to eliminate the arithmetic progression part. $(2) - (1):$ $3S - S = (1 \cdot 3^1 + 2 \cdot 3^2 + \dots + 10 \cdot 3^{10}) - (1 \cdot 3^0 + 2 \cdot 3^1 + \dots + 10 \cdot 3^9)$ Let's align the terms carefully for subtraction: $3S = \quad \quad 1 \cdot 3^1 + 2 \cdot 3^2 + \dots + 9 \cdot 3^9 + 10 \cdot 3^{10}$ $S = 1 \cdot 3^0 + 2 \cdot 3^1 + 3 \cdot 3^2 + \dots + 10 \cdot 3^9$ Subtracting term by term: $2S = -1 \cdot 3^0 + (1-2)3^1 + (2-3)3^2 + \dots + (9-10)3^9 + 10 \cdot 3^{10}$

Step 4: Simplify the Resulting Series $2S = -1 - 1 \cdot 3^1 - 1 \cdot 3^2 - \dots - 1 \cdot 3^9 + 10 \cdot 3^{10}$ Factor out $-1$ from the terms involving powers of 3: $2S = -(1 + 3^1 + 3^2 + \dots + 3^9) + 10 \cdot 3^{10}$ The terms in the parenthesis form a geometric progression.

Step 5: Sum the Geometric Progression The series inside the parenthesis is $1 + 3 + 3^2 + \dots + 3^9$. This is a GP with:

• First term ($A$) = 1
• Common ratio ($R$) = 3
• Number of terms ($N$) = 10 (from $3^0$ to $3^9$) Using the GP sum formula $S_N = \frac{A(R^N - 1)}{R-1}$: Sum of GP $= \frac{1(3^{10} - 1)}{3-1} = \frac{3^{10} - 1}{2}$.

Step 6: Substitute the GP Sum Back into the Equation for 2S Substitute the sum of the GP back into the equation from Step 4: $2S = - \left( \frac{3^{10} - 1}{2} \right) + 10 \cdot 3^{10}$ To simplify, find a common denominator: $2S = \frac{-(3^{10} - 1) + 2 \cdot (10 \cdot 3^{10})}{2}$ $2S = \frac{-3^{10} + 1 + 20 \cdot 3^{10}}{2}$ Combine the terms with $3^{10}$: $2S = \frac{(20-1) \cdot 3^{10} + 1}{2}$ $2S = \frac{19 \cdot 3^{10} + 1}{2}$

Step 7: Solve for S Divide both sides by 2 to find the value of $S$: $S = \frac{19 \cdot 3^{10} + 1}{4}$

Common Mistakes & Tips

• Incorrectly identifying $n$: Ensure the number of terms ($n$) is correctly counted for both the AP and GP parts. In this case, the AP terms are $1, 2, \dots, 10$, which gives $n=10$.
• Errors in subtraction: Carefully align terms with the same power of the common ratio when subtracting the series. Pay attention to the first term of the original series and the last term of the multiplied series.
• Formula application: Double-check the GP sum formula and ensure it's applied correctly with the right values for $A$, $R$, and $N$.

Summary

The given series is an Arithmetico-Geometric Progression (AGP). The standard method to sum an AGP is to multiply the series by its common ratio ($r=3$) and subtract the original series. This process converts the series into a form where the arithmetic progression is eliminated, leaving a geometric progression that can be summed using the standard GP sum formula. After performing these steps and simplifying, the sum of the series is found to be $\frac{19 \cdot 3^{10} + 1}{4}$.

The final answer is $\boxed{\text{A}}$.