31
Given that $1^{2}+2^{2} +3^{2} + ... +20^{2}$ = 2870, the value of $(2^{2}+4^{2}+6^{2}+...... +40^{2})$ is :
- A11480
- B5740
- C28700
- D2870
Answer not set
Log in or create a free account to reply and vote.
0
CCommunity
11480 is the best suited
0
CCommunity
22 + 42 + 62 + .... + 402
= 22 (12 + 22 + 32 +.... + 202)
= 4 × 2870 = 11480
= 22 (12 + 22 + 32 +.... + 202)
= 4 × 2870 = 11480