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    Q&A - P3 Math

    Scheduled Pinned Locked Moved Primary 3
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    • corneyAmberC Offline
      corneyAmber
      last edited by

      Dharma:

      Hi Suz855,ks2me & small,

      Hope this may be of interest.

      1/2006 is the common factor of the series.

      1/2006( 1+3+5+….+ 2003+2005)

      1 \t\t= 1 (1x1)
      1+3\t\t= 4 (2x2)
      1+3+5\t\t = 9 (3x3)
      1+3+5+7 \t= 16(4x4)

      To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

      For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

      Now,
      1+3+5+7+….+2003+2005

      To find the number of terms (2005+1)/2 = 1003

      So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5
      Thanks Dharma, interesting pattern observation, learn something new again. šŸ˜„

      1 Reply Last reply Reply Quote 0
      • D Offline
        Dharma
        last edited by

        ks2me:
        Dharma:


        Hi Suz855,ks2me & small,

        Hope this may be of interest.

        1/2006 is the common factor of the series.

        1/2006( 1+3+5+….+ 2003+2005)

        1 \t\t= 1 (1x1)
        1+3\t\t= 4 (2x2)
        1+3+5\t\t = 9 (3x3)
        1+3+5+7 \t= 16(4x4)

        To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

        For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

        Now,
        1+3+5+7+….+2003+2005

        To find the number of terms (2005+1)/2 = 1003

        So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5

        Thanks Dharma, interesting pattern observation, learn something new again. šŸ˜„

        Hi ks2me

        This is slight variation to this sort of question.

        2/2006 + 4/2006 + 6/2006 + …+ 2004/2006 + 2006/2006

        2/2006(1+2+3+…+ 1002+1003)

        Triangular numbers (If you use a dot to represent the sum of every series you will form a triangle)
        1 = 1 =(1x2)/2
        1+2 = 3 = (3x4)/2
        1+2+3 = 6 =(3x4)/2
        1+2+3+4 = 10 = (4x5)/2
        1+2+3+4+5 = 15 = (5x6)/2
        .
        .
        .
        1+2+3+4+ ….+ (N-1)+N =N(N+1)/2

        2/2006(1+2+3+…+ 1002+1003)
        =2/2006 x 1003x1004/2
        = 502

        1 Reply Last reply Reply Quote 0
        • corneyAmberC Offline
          corneyAmber
          last edited by

          Great sharing Dharma! :salute:

          1 Reply Last reply Reply Quote 0
          • D Offline
            Dharma
            last edited by

            ks2me:
            Great sharing Dharma! :salute:

            Glad u like it

            1 Reply Last reply Reply Quote 0
            • corneyAmberC Offline
              corneyAmber
              last edited by

              Dharma:
              ks2me:

              Great sharing Dharma! :salute:


              Glad u like it

              Yes I do but I wonder if P3/P4 children can appreciate it? My child is not at that level yet so cannot test on her.

              1 Reply Last reply Reply Quote 0
              • D Offline
                Dharma
                last edited by

                ks2me:
                Dharma:

                [quote=\"ks2me\"]Great sharing Dharma! :salute:


                Glad u like it

                Yes I do but I wonder if P3/P4 children can appreciate it? My child is not at that level yet so cannot test on her.[/quote]Those in P4 doing Olympiad Maths will be able to appreciate.

                Show them square numbers (series with consecutive odd nos) and triangular numbers in pictorial form. Will understand better after watching this http://www.yteach.co.uk/index.php/resources/triangular_square_numbers_root_t_page_3.html

                1 Reply Last reply Reply Quote 0
                • M Offline
                  Muffins
                  last edited by

                  Dharma:
                  Hi Suz855,ks2me & small,


                  Hope this may be of interest.

                  1/2006 is the common factor of the series.

                  1/2006( 1+3+5+….+ 2003+2005)

                  1 \t\t= 1 (1x1)
                  1+3\t\t= 4 (2x2)
                  1+3+5\t\t = 9 (3x3)
                  1+3+5+7 \t= 16(4x4)

                  To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

                  For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

                  Now,
                  1+3+5+7+….+2003+2005

                  To find the number of terms (2005+1)/2 = 1003

                  So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5
                  Thanks for the crystal clear explaining, buddy! :thankyou: :salute:

                  1 Reply Last reply Reply Quote 0
                  • S Offline
                    small
                    last edited by

                    [quote]Hi Suz855,ks2me & small,


                    Hope this may be of interest.

                    1/2006 is the common factor of the series.

                    1/2006( 1+3+5+….+ 2003+2005)

                    1 \t\t= 1 (1x1)
                    1+3\t\t= 4 (2x2)
                    1+3+5\t\t = 9 (3x3)
                    1+3+5+7 \t= 16(4x4)

                    To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

                    For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

                    Now,
                    1+3+5+7+….+2003+2005

                    To find the number of terms (2005+1)/2 = 1003

                    So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5[/quote]Hi Dharma,

                    Thanks for sharing...learned something new... :celebrate: :udaman:

                    1 Reply Last reply Reply Quote 0
                    • S Offline
                      sexymama
                      last edited by

                      Pls help me with my son's H/W... 😢


                      Minghua gave 1/2 of his money to Lily,.
                      In return, Lily gave 1/3 of whatever money
                      she had to Minghua.
                      Later, Minghua gave 1/4 of whatever
                      money he had then to Lily.
                      Finally, Minghua had $675 and Lily had $1325.
                      how much money di Minghua have at first?

                      1 Reply Last reply Reply Quote 0
                      • M Offline
                        mgsim
                        last edited by

                        is the answer $700 ?

                        1 Reply Last reply Reply Quote 0

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