Key Concepts and Formulas

• Sum of squares of the first n natural numbers: $\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$
• Sum of squares of the first n odd numbers: $\sum_{k=1}^{n} (2k-1)^2 = 1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}$
• Sum of squares of the first n even numbers: $\sum_{k=1}^{n} (2k)^2 = 2^2 + 4^2 + 6^2 + \dots + (2n)^2 = 4 \sum_{k=1}^{n} k^2 = 4 \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$

Step-by-Step Solution

Step 1: Analyze the given series and define A and B. The given series is $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$. We observe that the terms at odd positions are squares of odd numbers, i.e., $(2k-1)^2$, and the terms at even positions are twice the square of even numbers, i.e., $2 \cdot (2k)^2$.

Let $T_n$ be the $n$-th term of the series. If $n$ is odd, $n = 2k-1$, then $T_n = (2k-1)^2$. If $n$ is even, $n = 2k$, then $T_n = 2 \cdot (2k)^2$.

• A is the sum of the first 20 terms. This means there are 10 odd-positioned terms and 10 even-positioned terms.
• B is the sum of the first 40 terms. This means there are 20 odd-positioned terms and 20 even-positioned terms.

Step 2: Calculate A, the sum of the first 20 terms. The first 20 terms consist of odd-indexed terms up to $19$ and even-indexed terms up to $20$. The odd-indexed terms are $1^2, 3^2, 5^2, \dots, 19^2$. There are $(19-1)/2 + 1 = 10$ such terms. The even-indexed terms are $2 \cdot 2^2, 2 \cdot 4^2, 2 \cdot 6^2, \dots, 2 \cdot 20^2$. There are $20/2 = 10$ such terms.

So, $A = (1^2 + 3^2 + 5^2 + \dots + 19^2) + (2 \cdot 2^2 + 2 \cdot 4^2 + 2 \cdot 6^2 + \dots + 2 \cdot 20^2)$.

Let's calculate the sum of odd squares: $1^2 + 3^2 + \dots + 19^2$. Here $n=10$ for the formula $\sum_{k=1}^{n} (2k-1)^2$. Sum of odd squares = $\frac{10(2 \cdot 10 - 1)(2 \cdot 10 + 1)}{3} = \frac{10 \cdot 19 \cdot 21}{3} = 10 \cdot 19 \cdot 7 = 1330$.

Let's calculate the sum of even terms: $2(2^2 + 4^2 + 6^2 + \dots + 20^2)$. We can factor out 2: $2 \cdot 2^2 (1^2 + 2^2 + 3^2 + \dots + 10^2) = 8 (1^2 + 2^2 + 3^2 + \dots + 10^2)$. Using the sum of squares formula for $n=10$: $8 \cdot \frac{10(10+1)(2 \cdot 10 + 1)}{6} = 8 \cdot \frac{10 \cdot 11 \cdot 21}{6} = 8 \cdot \frac{2310}{6} = 8 \cdot 385 = 3080$.

Therefore, $A = 1330 + 3080 = 4410$.

Step 3: Calculate B, the sum of the first 40 terms. The first 40 terms consist of odd-indexed terms up to $39$ and even-indexed terms up to $40$. The odd-indexed terms are $1^2, 3^2, 5^2, \dots, 39^2$. There are $(39-1)/2 + 1 = 20$ such terms. The even-indexed terms are $2 \cdot 2^2, 2 \cdot 4^2, 2 \cdot 6^2, \dots, 2 \cdot 40^2$. There are $40/2 = 20$ such terms.

So, $B = (1^2 + 3^2 + 5^2 + \dots + 39^2) + (2 \cdot 2^2 + 2 \cdot 4^2 + 2 \cdot 6^2 + \dots + 2 \cdot 40^2)$.

Calculate the sum of odd squares: $1^2 + 3^2 + \dots + 39^2$. Here $n=20$ for the formula $\sum_{k=1}^{n} (2k-1)^2$. Sum of odd squares = $\frac{20(2 \cdot 20 - 1)(2 \cdot 20 + 1)}{3} = \frac{20 \cdot 39 \cdot 41}{3} = 20 \cdot 13 \cdot 41 = 10660$.

Calculate the sum of even terms: $2(2^2 + 4^2 + 6^2 + \dots + 40^2)$. Factor out 2: $2 \cdot 2^2 (1^2 + 2^2 + 3^2 + \dots + 20^2) = 8 (1^2 + 2^2 + 3^2 + \dots + 20^2)$. Using the sum of squares formula for $n=20$: $8 \cdot \frac{20(20+1)(2 \cdot 20 + 1)}{6} = 8 \cdot \frac{20 \cdot 21 \cdot 41}{6} = 8 \cdot \frac{17220}{6} = 8 \cdot 2870 = 22960$.

Therefore, $B = 10660 + 22960 = 33620$.

Step 4: Calculate B - 2A. We have $A = 4410$ and $B = 33620$. $B - 2A = 33620 - 2 \cdot 4410 = 33620 - 8820 = 24800$.

Step 5: Find the value of λ. We are given that $B - 2A = 100\lambda$. Substituting the value of $B - 2A$: $24800 = 100\lambda$. $\lambda = \frac{24800}{100} = 248$.

Common Mistakes and Tips

• Correctly identifying the number of terms: When calculating the sum of the first $N$ terms, be careful to determine how many odd and even numbers are involved. For $N$ terms, there are $N/2$ odd numbers and $N/2$ even numbers if $N$ is even. If $N$ is odd, there are $(N+1)/2$ odd numbers and $(N-1)/2$ even numbers.
• Using the correct formulas: Ensure you are using the formulas for the sum of squares of natural numbers and odd numbers accurately.
• Algebraic simplification: Simplify expressions like $2 \cdot (2k)^2$ to $8k^2$ before applying summation formulas to reduce errors.

Summary The problem involves calculating the sum of a series with a specific pattern for odd and even terms. We first identified the pattern and separated the series into sums of odd squares and scaled sums of even squares. By applying the standard formulas for the sum of squares of natural and odd numbers, we computed the values of A (sum of the first 20 terms) and B (sum of the first 40 terms). Finally, we used the given relation $B - 2A = 100\lambda$ to find the value of $\lambda$.

The final answer is \boxed{248}.