Key Concepts and Formulas

• Sum of the first $k$ natural numbers: The sum of the first $k$ positive integers is given by: $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$
• Sum of the first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
• Sum of the squares of the first $n$ natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Linearity of Summation: This property allows us to manipulate sums: $\sum_{k=1}^{n} (af(k) + bg(k)) = a \sum_{k=1}^{n} f(k) + b \sum_{k=1}^{n} g(k)$

Step-by-Step Solution

Step 1: Simplify the Inner Summation

• What we do: We first simplify the expression inside the parenthesis, which represents the sum of the first $k$ natural numbers.
• Why we do it: This transforms the general term of the outer summation into a compact algebraic expression, making it amenable to standard summation formulas. The sum $1 + 2 + 3 + \dots + k$ is an arithmetic progression. Using the formula for the sum of the first $k$ natural numbers: $1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2}$

Step 2: Rewrite the Overall Summation

• What we do: We substitute the simplified expression from Step 1 back into the original summation.
• Why we do it: This allows us to express the entire problem as a single summation of a polynomial in $k$. The given sum is: $S = \sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)}$ Substituting the simplified term, we get: $S = \sum\limits_{k = 1}^{20} {\frac{k(k+1)}{2}}$

Step 3: Expand and Apply Linearity

• What we do: We expand the term $\frac{k(k+1)}{2}$ and then use the linearity property of summation to split the sum into simpler components.
• Why we do it: Expanding $k(k+1)$ to $k^2+k$ allows us to use the formulas for the sum of squares and the sum of natural numbers. Factoring out constants and splitting the sum simplifies the calculation. $S = \sum\limits_{k = 1}^{20} {\frac{k^2 + k}{2}}$ Using the linearity property and factoring out the constant $\frac{1}{2}$: $S = \frac{1}{2} \sum\limits_{k = 1}^{20} {(k^2 + k)}$ $S = \frac{1}{2} \left( \sum\limits_{k = 1}^{20} {k^2} + \sum\limits_{k = 1}^{20} {k} \right)$

Step 4: Evaluate Standard Summations

• What we do: We now evaluate the two standard summations: $\sum_{k=1}^{20} k^2$ and $\sum_{k=1}^{20} k$.
• Why we do it: This is a direct application of the fundamental summation formulas with $n=20$. For the sum of the first 20 natural numbers: $\sum\limits_{k = 1}^{20} {k} = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210$ For the sum of the squares of the first 20 natural numbers: $\sum\limits_{k = 1}^{20} {k^2} = \frac{20(20+1)(2 \times 20+1)}{6} = \frac{20 \times 21 \times 41}{6}$ $\sum\limits_{k = 1}^{20} {k^2} = \frac{(2 \times 10) \times (3 \times 7) \times 41}{2 \times 3} = 10 \times 7 \times 41 = 70 \times 41 = 2870$

Step 5: Combine Results for the Final Answer

• What we do: We substitute the calculated values from Step 4 back into the expression for $S$ from Step 3 and perform the final arithmetic.
• Why we do it: This step brings together all the evaluated components to produce the final numerical answer to the problem. $S = \frac{1}{2} \left( 2870 + 210 \right)$ $S = \frac{1}{2} (3080)$ $S = 1540$

Common Mistakes & Tips

• Incorrectly Applying Formulas: Ensure you are using the correct formula for the sum of $k$ and the sum of $k^2$.
• Arithmetic Errors: Double-check all calculations, especially multiplication and division, as these are common sources of error.
• Simplification Order: Always simplify the inner summation first before dealing with the outer summation.

Summary

The problem requires evaluating a nested summation. We began by simplifying the inner sum, which is the sum of the first $k$ natural numbers, to $\frac{k(k+1)}{2}$. This transformed the original sum into $\sum_{k=1}^{20} \frac{k(k+1)}{2}$. By expanding the term and using the linearity of summation, we split this into the sum of squares and the sum of natural numbers. Applying the standard formulas for these sums with $n=20$, we calculated $\sum_{k=1}^{20} k^2 = 2870$ and $\sum_{k=1}^{20} k = 210$. Finally, we combined these results to find the total sum $S = \frac{1}{2}(2870 + 210) = 1540$.

The final answer is $\boxed{1540}$.