Maths · Arithmetic and Geometric progressions
=? \)
\( \left(2^{2}+4^{2}+6^{2}+\ldots \ldots+20^{2}\right)=? \)
- A. 77
- B. 1155
- C. 1540 \)
- D. \( 385 \times 385 \)
Step-by-step solution
The series is the sum of squares of the first 10 even numbers: 2^2 + 4^2 + ... + 20^2. Using the formula for sum of squares of first n even numbers: sum = 4 * (1^2 + 2^2 + ... + n^2) = 4 * n(n+1)(2n+1)/6 = (2n(n+1)(2n+1))/3. Substituting n=10 gives (2*10*11*21)/3 = 1540.
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