What is the value of (2^2 + 4^2 + 6^2 + ... + 20^2 )?
A) 542 B) 756 C) 1540 D) 968

• if we take 2^2 common of the sequence it reduces to 2^2(1^2+2^2+3^2+......+10^2) and now using the formula 1^2+2^2+3^2+........+n^2=n(n+1)(2n+1)/6 we get 2^2(10*11*21)/6=1540
• 8 years agoHelpfull: Yes(14) No(0)
• 2^2(1^1+2^2+3^3+.....+10^2)=1540
• 8 years agoHelpfull: Yes(4) No(1)
• first of all take 2^2 as common in all terms than the given expression can be rewritten as
  2^2(1^2 + 2^2 + 3^2 + . . . + 10^2)
  now as we know that (1^2 + 2^2 + 3^2 + . . . + n^2) = n(n+1)(2n+1)/6
  
  so from this we can solve this...
  2^2(10)(10+1)(2*10+1)/6 = 4*10*11*21/6 = 1540 ans.
• 8 years agoHelpfull: Yes(2) No(0)
• 1540
  4+16+36+64+100+144+196+256+324+400
  1540
• 8 years agoHelpfull: Yes(2) No(1)
• 2^2+2^2*2^2+2^2*3^2+2^2*4^2+..............2^2*10^2
  2^2(1^2+2^2+3^2+......10^2)
  4*5*11*7=1540
• 8 years agoHelpfull: Yes(1) No(0)
• sum of even number square = 2n(n+1)(2n+1)/3
  = 2*10(10+1)(2*10+1)/3
  =1540
• 8 years agoHelpfull: Yes(0) No(0)
• Take 4 as common from series, u will get,
  4(1^2+2^2+3^2+.....+10^2)
  Then, use formula,
  n(n+1)(2n+1)/6
  Ans- 1540
• 6 years agoHelpfull: Yes(0) No(0)