Given that $(1^2 + 2^2 + 3^2 + .... + 20^2) = 2870$, the value of $(2^2 + 4^2 + 6^2 + .... + 40^2)$ is

### Concept & Logic When every term in a series is multiplied by a constant, the sum of the series is also multiplied by that constant. In a series of squares, factoring out a common base requires squaring that common factor. $$ (ab)^2 = a^2 \times b^2 $$ ### Step-by-Step Solution * **Given:** $1^2 + 2^2 + 3^2 + .... + 20^2 = 2870$ Target Series: $2^2 + 4^2 + 6^2 + .... + 40^2$ * **Calculation / Deduction:** Notice that every term in the target series is an even number squared. We can rewrite each term as a multiple of $2$: $$ (2 \times 1)^2 + (2 \times 2)^2 + (2 \times 3)^2 + .... + (2 \times 20)^2 $$ Apply the exponent rule $(ab)^2 = a^2 \times b^2$: $$ (2^2 \times 1^2) + (2^2 \times 2^2) + (2^2 \times 3^2) + .... + (2^2 \times 20^2) $$ Factor out the common term $2^2$ (which is $4$): $$ 4 \times (1^2 + 2^2 + 3^2 + .... + 20^2) $$ Substitute the known value of the original series ($2870$): $$ 4 \times 2870 $$ $$ = 11480 $$ ### Exam Strategy & Shortcut **Ratio Method:** Recognize immediately that the second series represents the square of $2$ times the first series. The multiplier isn't $2$, it is $2^2 = 4$. Simply multiply the given total by $4$. $$ 2870 \times 4 = 11480 $$ ### Common Pitfall The most common mistake is assuming that doubling the base numbers simply doubles the final sum. Students often hastily calculate $2870 \times 2 = 5740$, failing to realize that squaring a doubled number quadruples its value. ### Final Answer **Therefore, the correct answer is 11480.**