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 $a, (a+d)r, (a+2d)r^2, \dots$.
• Sum of an AGP: The standard method involves multiplying the series by the common ratio ($r$) of the GP part and subtracting the new series from the original. This converts the AGP into a standard GP.
• Sum of a finite Geometric Progression (GP): $S_N = \frac{A(R^N - 1)}{R - 1}$, where $A$ is the first term, $R$ is the common ratio, and $N$ is the number of terms.
• Exponent Rules: $\frac{a^m}{a^n} = a^{m-n}$ and $a^m \cdot a^n = a^{m+n}$.

Step-by-Step Solution

Step 1: Identify the Series as an Arithmetico-Geometric Progression (AGP)

Let the given sum be $S$. $S = {(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + \dots + 10{(11)^9}$ We can rewrite the terms to identify the AP and GP components. The coefficients $1, 2, 3, \dots, 10$ form an Arithmetic Progression (AP) with first term $a=1$ and common difference $d=1$. The remaining parts of the terms involve powers of 10 and 11. Let's examine the ratio of consecutive terms: Term 1: $1 \cdot (10)^9$ Term 2: $2 \cdot (11)^1 \cdot (10)^8$ Term 3: $3 \cdot (11)^2 \cdot (10)^7$ The ratio of the GP part is $\frac{(11)^1 \cdot (10)^8}{(10)^9} = \frac{11}{10}$. This is the common ratio ($r$) of the Geometric Progression. The $i$-th term of the series can be written as $i \cdot (11)^{i-1} \cdot (10)^{10-i}$. For $i=1$, the term is $1 \cdot (11)^0 \cdot (10)^9 = (10)^9$. For $i=2$, the term is $2 \cdot (11)^1 \cdot (10)^8$. For $i=10$, the term is $10 \cdot (11)^9 \cdot (10)^0 = 10 \cdot (11)^9$. Thus, the series is an AGP with $a=1$, $d=1$, and $r = \frac{11}{10}$. The number of terms is $N=10$.

Step 2: Set up the Subtraction Method for AGP

Write out the series $S$: $S = 1 \cdot {(10)^9} + 2 \cdot {(11)^1}\,({10^8}) + 3 \cdot {(11)^2}\,({10^7}) + \dots + 9 \cdot {(11)^8}\,({10^1}) + 10 \cdot {(11)^9}$ Multiply $S$ by the common ratio $r = \frac{11}{10}$: $rS = \frac{11}{10}S = \frac{11}{10} \left[ 1 \cdot {(10)^9} + 2 \cdot {(11)^1}\,({10^8}) + \dots + 10 \cdot {(11)^9} \right]$ $rS = 1 \cdot \frac{11}{10}{(10)^9} + 2 \cdot \frac{11}{10}{(11)^1}\,({10^8}) + \dots + 10 \cdot \frac{11}{10}{(11)^9}$ Simplify each term in $rS$: $rS = 1 \cdot {(11)^1}\,({10^8}) + 2 \cdot {(11)^2}\,({10^7}) + \dots + 9 \cdot {(11)^9}\,({10^0}) + 10 \cdot {(11)^{10}}$ Now, align $S$ and $rS$ vertically and subtract $rS$ from $S$ to eliminate the arithmetic progression part.

$S \quad = \quad 1 \cdot {(10)^9} + 2 \cdot {(11)^1}\,({10^8}) + 3 \cdot {(11)^2}\,({10^7}) + \dots + 10 \cdot {(11)^9}$ $rS = \quad \quad \quad 1 \cdot {(11)^1}\,({10^8}) + 2 \cdot {(11)^2}\,({10^7}) + \dots + 9 \cdot {(11)^9} + 10 \cdot {(11)^{10}}$

Step 3: Perform the Subtraction ($S - rS$)

Subtracting the two equations: $(1-r)S = S - rS = \left(1 - \frac{11}{10}\right)S = -\frac{1}{10}S$ $-\frac{1}{10}S = 1 \cdot {(10)^9} + (2-1){(11)^1}\,({10^8}) + (3-2){(11)^2}\,({10^7}) + \dots + (10-9){(11)^9} - 10 \cdot {(11)^{10}}$ The coefficients of the terms $(11)^{i-1}(10)^{10-i}$ for $i=1$ to $9$ become $1$ because the AP common difference is $1$. $-\frac{1}{10}S = {(10)^9} + {(11)^1}\,({10^8}) + {(11)^2}\,({10^7}) + \dots + {(11)^9} - 10 \cdot {(11)^{10}}$

Step 4: Sum the Resulting Geometric Progression

The terms $G_{sum} = {(10)^9} + {(11)^1}\,({10^8}) + {(11)^2}\,({10^7}) + \dots + {(11)^9}$ form a Geometric Progression. The first term is $A = {(10)^9}$. The common ratio is $R = \frac{(11)^1\,({10^8})}{{(10)^9}} = \frac{11}{10}$. The number of terms is $N=10$ (powers of 11 from 0 to 9).

Using the GP sum formula $S_N = \frac{A(R^N - 1)}{R - 1}$: $G_{sum} = \frac{{(10)^9} \left( \left(\frac{11}{10}\right)^{10} - 1 \right)}{\frac{11}{10} - 1}$ $G_{sum} = \frac{{(10)^9} \left( \frac{11^{10}}{10^{10}} - 1 \right)}{\frac{1}{10}}$ $G_{sum} = 10 \cdot {(10)^9} \left( \frac{11^{10} - 10^{10}}{10^{10}} \right)$ $G_{sum} = \frac{10^{10}}{10^{10}} (11^{10} - 10^{10})$ $G_{sum} = 11^{10} - 10^{10}$

Step 5: Solve for S

Substitute the sum of the GP back into the equation from Step 3: $-\frac{1}{10}S = G_{sum} - 10 \cdot {(11)^{10}}$ $-\frac{1}{10}S = (11^{10} - 10^{10}) - 10 \cdot {(11)^{10}}$ $-\frac{1}{10}S = 11^{10} - 10^{10} - 10 \cdot 11^{10}$ $-\frac{1}{10}S = -9 \cdot 11^{10} - 10^{10}$ Correction: Let's re-examine the subtraction. $S = 1 \cdot {(10)^9} + 2 \cdot {(11)^1}\,({10^8}) + \dots + 10 \cdot {(11)^9}$ $rS = \quad \quad 1 \cdot {(11)^1}\,({10^8}) + \dots + 9 \cdot {(11)^9} + 10 \cdot {(11)^{10}}$ $(1-r)S = 1 \cdot {(10)^9} + 1 \cdot {(11)^1}\,({10^8}) + \dots + 1 \cdot {(11)^9} - 10 \cdot {(11)^{10}}$ The sum of the first 10 terms is $G_{sum} = {(10)^9} + {(11)^1}\,({10^8}) + \dots + {(11)^9} = 11^{10} - 10^{10}$. So, $-\frac{1}{10}S = (11^{10} - 10^{10}) - 10 \cdot {(11)^{10}}$ $-\frac{1}{10}S = 11^{10} - 10^{10} - 10 \cdot 11^{10}$ $-\frac{1}{10}S = -9 \cdot 11^{10} - 10^{10}$ There seems to be an error in the subtraction or the GP sum. Let's re-evaluate the GP sum.

The AGP is $S = \sum_{i=1}^{10} i \cdot (11)^{i-1} \cdot (10)^{10-i}$. Let $x = \frac{11}{10}$. The terms are $i \cdot x^{i-1} \cdot 10^{10-i}$. $S = 1 \cdot 10^9 + 2 \cdot 11 \cdot 10^8 + 3 \cdot 11^2 \cdot 10^7 + \dots + 10 \cdot 11^9 \cdot 10^0$. $S = 10^9 \left[ 1 + 2 \left(\frac{11}{10}\right) + 3 \left(\frac{11}{10}\right)^2 + \dots + 10 \left(\frac{11}{10}\right)^9 \right]$. Let $x = \frac{11}{10}$. $S = 10^9 \left[ 1 + 2x + 3x^2 + \dots + 10x^9 \right]$. Let $S' = 1 + 2x + 3x^2 + \dots + 10x^9$. $S' = \frac{d}{dx} (1 + x + x^2 + \dots + x^{10}) = \frac{d}{dx} \left(\frac{x^{11}-1}{x-1}\right)$. $S' = \frac{11x^{10}(x-1) - (x^{11}-1)}{(x-1)^2} = \frac{11x^{11} - 11x^{10} - x^{11} + 1}{(x-1)^2} = \frac{10x^{11} - 11x^{10} + 1}{(x-1)^2}$. With $x = \frac{11}{10}$: $x-1 = \frac{1}{10}$. $S' = \frac{10(\frac{11}{10})^{11} - 11(\frac{11}{10})^{10} + 1}{(\frac{1}{10})^2} = 100 \left[ \frac{10 \cdot 11^{11}}{10^{11}} - \frac{11 \cdot 11^{10}}{10^{10}} + 1 \right]$ $S' = 100 \left[ \frac{11^{11}}{10^{10}} - \frac{11^{11}}{10^{10}} + 1 \right] = 100 [1] = 100$. So, $S = 10^9 \cdot S' = 10^9 \cdot 100 = 10^{11}$.

Let's re-do the subtraction method carefully. $S = 1 \cdot {(10)^9} + 2 \cdot {(11)^1}\,({10^8}) + 3 \cdot {(11)^2}\,({10^7}) + \dots + 10 \cdot {(11)^9}$ $rS = \quad \quad 1 \cdot {(11)^1}\,({10^8}) + 2 \cdot {(11)^2}\,({10^7}) + \dots + 9 \cdot {(11)^9} + 10 \cdot {(11)^{10}}$ $(1-r)S = S - rS = 1 \cdot {(10)^9} + (2-1){(11)^1}\,({10^8}) + (3-2){(11)^2}\,({10^7}) + \dots + (10-9){(11)^9} - 10 \cdot {(11)^{10}}$ $(1-r)S = {(10)^9} + {(11)^1}\,({10^8}) + {(11)^2}\,({10^7}) + \dots + {(11)^9} - 10 \cdot {(11)^{10}}$ The sum of the first 10 terms is a GP: $A = 10^9$, $R = \frac{11}{10}$, $N=10$. Sum of this GP: $\frac{10^9 ((\frac{11}{10})^{10} - 1)}{\frac{11}{10} - 1} = \frac{10^9 (\frac{11^{10}}{10^{10}} - 1)}{\frac{1}{10}} = 10 \cdot 10^9 \frac{11^{10} - 10^{10}}{10^{10}} = \frac{10^{10}}{10^{10}}(11^{10} - 10^{10}) = 11^{10} - 10^{10}$. So, $-\frac{1}{10}S = (11^{10} - 10^{10}) - 10 \cdot {(11)^{10}}$ $-\frac{1}{10}S = 11^{10} - 10^{10} - 10 \cdot 11^{10}$ $-\frac{1}{10}S = -9 \cdot 11^{10} - 10^{10}$ This still leads to a complex result. Let's check the last term of $S$. The last term is $10 \cdot (11)^9$. The last term of $rS$ is $10 \cdot \frac{11}{10} \cdot (11)^9 = 11 \cdot 11^9 = 11^{10}$. The terms in $S-rS$ are: $1 \cdot 10^9$ (first term of S) $+ (2-1) \cdot 11^1 \cdot 10^8 = 1 \cdot 11^1 \cdot 10^8$ $+ (3-2) \cdot 11^2 \cdot 10^7 = 1 \cdot 11^2 \cdot 10^7$ ... $+ (10-9) \cdot 11^9 \cdot 10^0 = 1 \cdot 11^9$ $- 10 \cdot 11^{10}$ (last term of rS)

The sum of the GP part is $10^9 + 11 \cdot 10^8 + 11^2 \cdot 10^7 + \dots + 11^9$. This is a GP with first term $A = 10^9$, ratio $R = \frac{11}{10}$, and $N=10$ terms. Sum $= \frac{10^9 \left( (\frac{11}{10})^{10} - 1 \right)}{\frac{11}{10} - 1} = \frac{10^9 (\frac{11^{10}}{10^{10}} - 1)}{\frac{1}{10}} = 10 \cdot 10^9 \frac{11^{10} - 10^{10}}{10^{10}} = 11^{10} - 10^{10}$. So, $-\frac{1}{10}S = (11^{10} - 10^{10}) - 10 \cdot 11^{10}$ $-\frac{1}{10}S = 11^{10} - 10^{10} - 10 \cdot 11^{10} = -9 \cdot 11^{10} - 10^{10}$

Let's check the original problem statement and question. $S = k(10)^9$. We found $S = 10^{11}$ using the derivative method. Let's trust that for now and see if the subtraction method can be made to yield this.

Let's try rewriting the series $S$ slightly differently. $S = \sum_{i=1}^{10} i \cdot (11)^{i-1} \cdot (10)^{10-i}$ $S = 10^9 \sum_{i=1}^{10} i \left(\frac{11}{10}\right)^{i-1}$ Let $x = \frac{11}{10}$. $S = 10^9 \sum_{i=1}^{10} i x^{i-1}$ Let $T = \sum_{i=1}^{10} i x^{i-1} = 1 + 2x + 3x^2 + \dots + 10x^9$. $xT = x + 2x^2 + 3x^3 + \dots + 10x^{10}$. $T - xT = (1 + 2x + \dots + 10x^9) - (x + 2x^2 + \dots + 10x^{10})$ $(1-x)T = 1 + x + x^2 + \dots + x^9 - 10x^{10}$. The sum $1 + x + \dots + x^9$ is a GP with $A=1$, $R=x$, $N=10$. Sum $= \frac{1(x^{10}-1)}{x-1}$. $(1-x)T = \frac{x^{10}-1}{x-1} - 10x^{10}$. $T = \frac{1}{1-x} \left( \frac{x^{10}-1}{x-1} - 10x^{10} \right) = \frac{1}{-(x-1)} \left( \frac{x^{10}-1}{x-1} - 10x^{10} \right)$ $T = \frac{1-x^{10}}{(x-1)^2} + \frac{10x^{10}}{x-1}$.

Using $x = \frac{11}{10}$, $x-1 = \frac{1}{10}$, $1-x = -\frac{1}{10}$. $T = \frac{1 - (\frac{11}{10})^{10}}{(\frac{1}{10})^2} + \frac{10(\frac{11}{10})^{10}}{\frac{1}{10}}$ $T = 100 \left( 1 - \frac{11^{10}}{10^{10}} \right) + 100 \left(\frac{11}{10}\right)^{10}$ $T = 100 - 100 \frac{11^{10}}{10^{10}} + 100 \frac{11^{10}}{10^{10}} = 100$. This confirms $T=100$.

So $S = 10^9 \cdot T = 10^9 \cdot 100 = 10^{11}$.

Step 6: Determine k

We are given that $S = k{(10)^9}$. We found $S = 10^{11}$. Equating these two: $k{(10)^9} = 10^{11}$ Divide both sides by $(10)^9$: $k = \frac{10^{11}}{10^9}$ Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$: $k = 10^{11-9}$ $k = 10^2$ $k = 100$.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with exponent rules and fraction manipulations, especially during the subtraction and simplification of the geometric series.
• Off-by-One Errors: Ensure the number of terms in the AGP and the resulting GP are correctly identified.
• Misidentification of GP Ratio: The common ratio of the GP part must be correctly determined from consecutive terms.

Summary

The given sum is an Arithmetico-Geometric Progression. By applying the standard method of multiplying the series by its common ratio ($r = \frac{11}{10}$) and subtracting, we transformed the AGP into a sum involving a standard Geometric Progression. Alternatively, using calculus by differentiating a related geometric series sum provides a more direct path to the sum of the arithmetic coefficients part. Both methods confirm that the total sum $S$ is $10^{11}$. Equating this to the given form $k{(10)^9}$ allows us to solve for $k$, yielding $k=100$.

The final answer is $\boxed{100}$.