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Math paper4 Jun. 2009 help

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moon:
plz, can any body help me in math paper 4 Jun.09 question 11 d (i,ii) ??? ??? ???

wassup:
11(d)(i) (n(n+1))/2 + ((n+1)(n+2))/2
         =(n^2+n+n^2+2n+n+2)/2
         =(2n^2+4n+2)/2
         =(2(n^2+2n+1))/2
         =n^2+2n+1
         So, n^2+2n+1=(n+1)^2

       (ii) (n+1)^2=3481
             n^2+2n+1=3481
             n^2+2n-3480=0
             n=(-2+-underroot(2^2-4*1*-3480))/2
             n=(-2+-underroot(13924))/2
             n=(-2+-118)/2
             So, n=58, -60
             n=58 (It can't be negative)
             Two terms will be n, (n+1)
             =58, 59
             (58(58+1))/2=1711
             (59(59+1))/2=1770
             Answer=1711 and 1770

moon:
thanks a lot :) :)

wassup:
welcome

moon:

--- Quote from: wassup on October 20, 2009, 12:58:00 pm ---11(d)(i) (n(n+1))/2 + ((n+1)(n+2))/2
         =(n^2+n+n^2+2n+n+2)/2
         =(2n^2+4n+2)/2
         =(2(n^2+2n+1))/2
         =n^2+2n+1
         So, n^2+2n+1=(n+1)^2

       (ii) (n+1)^2=3481
             n^2+2n+1=3481
             n^2+2n-3480=0
             n=(-2+-underroot(2^2-4*1*-3480))/2
             n=(-2+-underroot(13924))/2
             n=(-2+-118)/2
             So, n=58, -60
             n=58 (It can't be negative)
             Two terms will be n, (n+1)
             =58, 59
             (58(58+1))/2=1711
             (59(59+1))/2=1770
             Answer=1711 and 1770

--- End quote ---

Please explain both parts still didn't get it ...

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