Results 1 to 3 of 3

Math Help - simple substitution in formula

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    Talking simple substitution in formula

    given that:
     1^2 + 2^2 + 3^2 +.... + n^2 = \frac{n(n+1)(2n+1)}{6}
    then evaluate  2^2 + 4^2 + 6^2 + ...... + 100^2

    looks easy... but i can't get it
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    2^2+4^2+6^2+........+100^2=[1(2)]^2+[2(2)]^2+[3(2)]^2+......+[50(2)]^2

    =[1^2+2^2+3^2+.......+50^2]2^2

    The number of terms is 50.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by differentiate View Post
    given that:
     1^2 + 2^2 + 3^2 +.... + n^2 = \frac{n(n+1)(2n+1)}{6}
    then evaluate  2^2 + 4^2 + 6^2 + ...... + 100^2

    looks easy... but i can't get it
    HI

    another approach , of course Archie's one is better ,

     1^2 + 2^2 + 3^2 +.... + n^2 = \frac{n(n+1)(2n+1)}{6} , Put n=100 , the sum of this would be 338350

    then 1^2+3^2+5^2+...+(2n-1)^2=\sum^{n}_{k=1}(2k-1)^2
    \sum^{n}_{k=1}(4k^2-4k+1)=\frac{2n(n+1)(2n+1)}{3}-2n(n+1)+n

    put n=50 , and this sum is 166650

    take their difference then .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 22nd 2011, 01:31 PM
  2. Replies: 4
    Last Post: September 5th 2010, 06:27 PM
  3. simple substitution problem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 9th 2009, 08:37 AM
  4. Replies: 2
    Last Post: May 2nd 2009, 10:14 PM
  5. Replies: 1
    Last Post: November 1st 2006, 08:45 AM

Search Tags


/mathhelpforum @mathhelpforum