Key Concepts and Formulas

• Summation Notation: The ability to express a sum of terms using sigma notation ($\sum$).
• Summation Formulas: The standard formulas for the sum of the first $n$ natural numbers and the sum of the squares of the first $n$ natural numbers:
  • $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
  • $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Algebraic Simplification: Using algebraic identities to simplify expressions, particularly involving squares of binomials.

Step-by-Step Solution

Step 1: Analyze the Series and Group Terms The given series is $2 \cdot 2^{2}-3^{2}+2 \cdot 4^{2}-5^{2}+2 \cdot 6^{2}-\ldots \ldots$. We are asked to find the sum of the first 20 terms. Observing the pattern, we can see that terms appear in pairs: $(2 \cdot 2^2 - 3^2)$, $(2 \cdot 4^2 - 5^2)$, $(2 \cdot 6^2 - 7^2)$, and so on. Since we need the sum of 20 terms and each group consists of 2 terms, there will be $20/2 = 10$ such groups. This means we need to find the sum of these 10 groups.

Step 2: Determine the General Term of a Group Let's find a general expression for the $r$-th group. The first term in each group involves $2 \cdot (\text{even number})^2$, and the second term is $-(\text{the next odd number})^2$. For the 1st group ($r=1$): $2 \cdot 2^2 - 3^2$. Here, the even number is $2 = 2 \cdot 1$, and the odd number is $3 = 2 \cdot 1 + 1$. For the 2nd group ($r=2$): $2 \cdot 4^2 - 5^2$. Here, the even number is $4 = 2 \cdot 2$, and the odd number is $5 = 2 \cdot 2 + 1$. For the 3rd group ($r=3$): $2 \cdot 6^2 - 7^2$. Here, the even number is $6 = 2 \cdot 3$, and the odd number is $7 = 2 \cdot 3 + 1$. Following this pattern, the $r$-th group, let's call it $T_r$, can be written as: $T_r = 2 \cdot (2r)^2 - (2r+1)^2$

Step 3: Simplify the General Term We simplify the expression for $T_r$ using algebraic identities: $T_r = 2 \cdot (2r)^2 - (2r+1)^2$ $T_r = 2 \cdot (4r^2) - ((2r)^2 + 2(2r)(1) + 1^2)$ $T_r = 8r^2 - (4r^2 + 4r + 1)$ Distributing the negative sign: $T_r = 8r^2 - 4r^2 - 4r - 1$ Combining like terms: $T_r = 4r^2 - 4r - 1$

Step 4: Set up the Summation We need to find the sum of these 10 groups, from $r=1$ to $r=10$. The total sum, $S_{20}$, is: $S_{20} = \sum_{r=1}^{10} T_r = \sum_{r=1}^{10} (4r^2 - 4r - 1)$ Using the linearity property of summation: $S_{20} = 4 \sum_{r=1}^{10} r^2 - 4 \sum_{r=1}^{10} r - \sum_{r=1}^{10} 1$

Step 5: Apply Summation Formulas and Calculate We use the standard summation formulas with $n=10$:

• $\sum_{r=1}^{10} r = \frac{10(10+1)}{2} = \frac{10 \cdot 11}{2} = 55$
• $\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385$
• $\sum_{r=1}^{10} 1 = 10 \cdot 1 = 10$

Substitute these values back into the expression for $S_{20}$: $S_{20} = 4(385) - 4(55) - 10$ Now, we perform the arithmetic: $S_{20} = 1540 - 220 - 10$ $S_{20} = 1540 - 230$ $S_{20} = 1310$

Common Mistakes & Tips

• Incorrect Number of Terms for Summation: A common mistake is to use $n=20$ in the summation formulas instead of $n=10$. Since the general term $T_r$ represents a pair of terms from the original series, we sum 10 such general terms.
• Algebraic Errors: Be careful when expanding squares of binomials and distributing negative signs. Double-checking these steps can prevent errors.
• Simplification of Fractions: Simplifying fractions like $\frac{10 \cdot 11 \cdot 21}{6}$ before multiplying can make calculations easier and less prone to errors. For example, $\frac{10 \cdot 11 \cdot 21}{6} = \frac{10}{2} \cdot 11 \cdot \frac{21}{3} = 5 \cdot 11 \cdot 7 = 385$.

Summary The problem requires summing a series by first identifying a pattern of grouped terms. By grouping the series into pairs and deriving a general term for each pair, we transformed the problem into summing a quadratic expression over 10 terms. Applying standard summation formulas for $r$ and $r^2$, and performing careful algebraic simplification and arithmetic, we arrived at the total sum.

The final answer is \boxed{1310}.