Key Concepts and Formulas

• Arithmetico-Geometric Progression (AGP): A series where each term is a product of a term from an Arithmetic Progression (AP) and a term from a Geometric Progression (GP). The general form is $a, (a+d)r, (a+2d)r^2, \ldots$.
• Sum of an AGP: The sum of an AGP can be found by multiplying the series by the common ratio $r$ of the GP part and subtracting the result from the original series.
• Sum of a finite Geometric Progression (GP): For a GP with first term $A$, common ratio $r$ ($r \neq 1$), and $N$ terms, the sum is $S_N = \frac{A(1-r^N)}{1-r}$.

Step-by-Step Solution

Step 1: Identify the structure of the given series. The given series is $S = (20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}$. We can observe that this is an Arithmetico-Geometric Progression (AGP). The terms can be represented as $T_n = n \cdot (21)^{n-1} \cdot (20)^{20-n}$ for $n = 1, 2, \ldots, 20$. Let's verify for $n=1$: $T_1 = 1 \cdot (21)^{0} \cdot (20)^{19} = (20)^{19}$. For $n=2$: $T_2 = 2 \cdot (21)^{1} \cdot (20)^{18} = 2(21)(20)^{18}$. For $n=20$: $T_{20} = 20 \cdot (21)^{19} \cdot (20)^{0} = 20(21)^{19}$. The series has $N=20$ terms.

To simplify, we can factor out $(20)^{19}$ from each term: $S = (20)^{19} \left[ 1 \cdot \left(\frac{21}{20}\right)^0 + 2 \cdot \left(\frac{21}{20}\right)^1 + 3 \cdot \left(\frac{21}{20}\right)^2 + \ldots + 20 \cdot \left(\frac{21}{20}\right)^{19} \right]$ Let $r = \frac{21}{20}$ and let $S' = 1 + 2r + 3r^2 + \ldots + 20r^{19}$. Then $S = (20)^{19} S'$. Our goal is to find $S'$ and then determine $k$.

Step 2: Sum the Arithmetico-Geometric Progression $S'$. $S' = 1 + 2r + 3r^2 + \ldots + 19r^{18} + 20r^{19} \quad \ldots (1)$ Multiply $S'$ by the common ratio $r$: $rS' = r + 2r^2 + 3r^3 + \ldots + 19r^{19} + 20r^{20} \quad \ldots (2)$

Subtract equation (2) from equation (1): $(1-r)S' = (1-0) + (2r-r) + (3r^2-2r^2) + \ldots + (20r^{19}-19r^{19}) - 20r^{20}$ $(1-r)S' = 1 + r + r^2 + \ldots + r^{19} - 20r^{20}$

The terms $1 + r + r^2 + \ldots + r^{19}$ form a finite GP with first term $A_{GP}=1$, common ratio $r$, and $N=20$ terms. The sum of this GP is: $S_{GP} = \frac{1(1-r^{20})}{1-r} = \frac{1-r^{20}}{1-r}$

Substitute this sum back into the expression for $(1-r)S'$: $(1-r)S' = \frac{1-r^{20}}{1-r} - 20r^{20}$

Now, substitute the value of $r = \frac{21}{20}$: $1-r = 1 - \frac{21}{20} = -\frac{1}{20}$.

$(-\frac{1}{20})S' = \frac{1 - \left(\frac{21}{20}\right)^{20}}{-\frac{1}{20}} - 20 \left(\frac{21}{20}\right)^{20}$ Multiply both sides by $-20$: $S' = -20 \left( \frac{1 - \left(\frac{21}{20}\right)^{20}}{-\frac{1}{20}} \right) - 20 \left( -20 \left(\frac{21}{20}\right)^{20} \right)$ $S' = -20 \left( -20 \left(1 - \left(\frac{21}{20}\right)^{20}\right) \right) + 400 \left(\frac{21}{20}\right)^{20}$ $S' = 400 \left(1 - \left(\frac{21}{20}\right)^{20}\right) + 400 \left(\frac{21}{20}\right)^{20}$ $S' = 400 - 400 \left(\frac{21}{20}\right)^{20} + 400 \left(\frac{21}{20}\right)^{20}$ $S' = 400$

Step 3: Calculate the value of $k$. We have $S = (20)^{19} S'$ and we are given $S = k(20)^{19}$. Substituting the value of $S'$: $S = (20)^{19} \cdot 400$ Comparing this with the given equation $S = k(20)^{19}$: $k(20)^{19} = 400(20)^{19}$ Dividing both sides by $(20)^{19}$: $k = 400$

Common Mistakes & Tips

• Algebraic Errors: Be very careful with signs and fractions when subtracting the series multiplied by $r$. A small mistake can lead to a completely wrong answer.
• Identifying $r$ and $N$: Correctly identifying the common ratio $r$ of the GP part and the number of terms $N$ is crucial for applying the AGP summation formula. In this case, $r = 21/20$ and $N = 20$.
• Formula Application: Ensure you are using the correct formula for the sum of a finite GP. The sum of $1+r+\dots+r^{N-1}$ is $\frac{1-r^N}{1-r}$.

Summary

The problem involves summing an Arithmetico-Geometric Progression. By recognizing the pattern of the series and factoring out a common term, we reduced it to summing a standard AGP of the form $1 + 2x + 3x^2 + \ldots + Nx^{N-1}$. The standard method of multiplying by the common ratio $r$ and subtracting the series was applied. This process isolated a finite geometric series, whose sum was then calculated. Finally, the sum of the AGP was used to find the value of $k$.

The final answer is $\boxed{400}$.