Given that $(1^2 + 2^2 + 3^2 + ..... + 10^2) = 385$, the value of $(2^2 + 4^2 + 6^2 + ..... + 20^2)$ is equal to

### Concept & Logic This problem tests your ability to recognize algebraic patterns and factor out common multipliers from a series. Instead of calculating the squares manually, look for the relationship between the bases in the two series. Notice that every term in the second series is exactly double the base of the corresponding term in the first series. ### Step-by-Step Solution * **Step 1:** Write out the target series. Target $= (2^2 + 4^2 + 6^2 + ..... + 20^2)$ * **Step 2:** Express each base as a multiple of the terms from the given series. $2^2 = (2 \times 1)^2 = 2^2 \times 1^2$ $4^2 = (2 \times 2)^2 = 2^2 \times 2^2$ $6^2 = (2 \times 3)^2 = 2^2 \times 3^2$ ... $20^2 = (2 \times 10)^2 = 2^2 \times 10^2$ * **Step 3:** Substitute these back into the target series and factor out the common term. Target $= (2^2 \times 1^2) + (2^2 \times 2^2) + (2^2 \times 3^2) + ..... + (2^2 \times 10^2)$ Target $= 2^2(1^2 + 2^2 + 3^2 + ..... + 10^2)$ * **Step 4:** Substitute the given value for the sum of squares. We are given that $(1^2 + 2^2 + 3^2 + ..... + 10^2) = 385$. Target $= 4 \times 385$ * **Step 5:** Calculate the final product. $4 \times 385 = 1540$ ### Exam Strategy & Shortcut Whenever you see a series of even squares being compared to a series of consecutive integer squares, remember the multiplier is always $2^2 = 4$. You do not need to write out the steps. Just mentally spot that $2^2$ factors out, leaving the original series. Immediately multiply the given sum by 4: $385 \times 4 = 1540$. ### Common Pitfall The most common trap is factoring out just $2$ instead of $2^2$. Students often incorrectly assume that if the bases are doubled, the sum is doubled, leading them to choose $770$. Because the terms are squared, the factor factored out must also be squared! ### Final Answer **Therefore, the correct answer is 1540.**