(11^2 + 12^2 + 13^2 + ... + 20^2) = ?

• The sum of squares of first n natural numbers is:
  
  (n*(n+1)*(2n+1))/6
  
  Thus find out the sum of squares of first 20 natural numbers(1-20) i.e 2870
  and now the sum squares of first 10 natural numbers(1-10) is : 385
  
  and the difference between these two will give you the sum of (11^2 + 12^2 + 13^2 + ... + 20^2) which is equal to 2485
• 11 years agoHelpfull: Yes(45) No(0)
• Okay So here we go:
  
  Re-write this as
  (10+1)^2+(10+2)^2+(10+3)^2...+(10+10)^2
  This would give us:
  
  10*10^2 + (1^2+2^2+3^2+..+10^2) + (20+40+60+..200)
  
  So now you get:
  1000+Sum of square 10 natural no.+ Sum of an A.P for 10 terms
  
  Answer:2485
• 11 years agoHelpfull: Yes(12) No(4)
• (11^2 + 12^2 + 13^2 + ... + 20^2) = (1^2+2^2+.............20^2)-(1^2+2^2+.........10^2)
  {(20*21*41)/6-(10*11*12)/6}=(2870-385)=2485
  
  
  
• 11 years agoHelpfull: Yes(6) No(2)
• first of all we should be clear of square numbers..
  11^2=121 12^2=144 13^2=169 14^2=196 15^2=225 16^2=256 17^2=289 18^2=324 19^2=361 20^=400
  121+144+169+196+225+256+289+324+361+400=2485
  this is easy to find square of numbers: for eg: xy^2= you must first square the ending digit y and if it comes as double digit number then we must add that to+ xy+y
  eg: 14^2
  4*4=16
  hence keep that 6 as it is
  now 14+4+1=19 hence the answer is 196
  
• 11 years agoHelpfull: Yes(2) No(4)
• 121+144+169+196+225+256+289+324+361{as i hav learnt da squares}
  =2085{added horizontally rather dan vertically writing inorder to save time}
• 11 years agoHelpfull: Yes(0) No(15)
• (10+1)^2+(10+2)^2+(10+3)^2+....+(10+10)^2
  10^2+1^2+2*10*1+10^2+2^2+2*10*2+....+10^2+10^2+2*10*10
  (10^2*10)+(1^2+2^2+3^2+...10^2)+2*10*(1+2+3+...+10)
  sum of n natural no=(n(n+1))/2
  sum of squares of n natural nos=n(n+1)(2n+1)/6
  here n=10
  1000+385+55=1440
• 11 years agoHelpfull: Yes(0) No(2)
• value=2485
• 11 years agoHelpfull: Yes(0) No(0)