Find the sum of the series \[({2^2} + {4^2} + {6^2} + ... + {20^2})\] .

Hint:- Sum of first n square terms is equal to $\dfrac{{{\text{n(n + 1)(2n + 1)}}}}{6}$.

Given, a series \[({2^2} + {4^2} + {6^2} + ... + {20^2})\]. We need to find the sum.
So, observing the series we can see that it is the sum of even squares only. That means 2 is common multiple of all. By simplifying the series, we get
$({2^2} + {4^2} + {6^2} + ... + {20^2}) = \left\{ {{{\left( {1 \times 2} \right)}^2} + {{\left( {2 \times 2} \right)}^2} + {{\left( {3 \times 2} \right)}^2} + ... + {{(2 \times 10)}^2}} \right\}$
And on further solving , we get
$({2^2} + {4^2} + {6^2} + ... + {20^2}) = {2^2}\left\{ {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + ... + {{\left( {10} \right)}^2}} \right\}$ -(1)
Sum of first n square terms is equal to $\dfrac{{{\text{n(n + 1)(2n + 1)}}}}{6}$
For equation (1) , we have n = 10
Applying the formula we get,
$
   \Rightarrow ({1^2} + {2^2} + {3^2} + ... + {10^2}) = \dfrac{{10(10 + 1)(2 \times 10 + 1)}}{6} \\
   \Rightarrow ({1^2} + {2^2} + {3^2} + ... + {10^2}) = \dfrac{{10 \times 11 \times 21}}{6} = 385 \\
 $
Putting the sum in equation (1), we get
$
   \Rightarrow ({2^2} + {4^2} + {6^2} + ... + {20^2}) = {2^2} \times 385 \\
   \Rightarrow ({2^2} + {4^2} + {6^2} + ... + {20^2}) = 1540 \\
 $
The sum of the given series is 1540.
Note:- In these types of questions, we need to understand a pattern. The series can be of few types only i.e. arithmetic , geometric, harmonic, Fibonacci ,sum of squares and sum of multiples of consecutive number (n*(n+1)).Other series are derived from these fundamentals.