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Re: Number System Problems

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  • coolharsha_b
    Here goes the soln 1)It can be seen that any term in the series can be written as Tn=(2n+1)/(n(n+1))^2 2n+1=(n+1)^2-n^2 Tn=n^-2 - (n+1)^-2 Summation to n terms
    Message 1 of 3 , Jul 30, 2003
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      Here goes the soln
      1)It can be seen that any term in the series can be written as
      Tn=(2n+1)/(n(n+1))^2
      2n+1=(n+1)^2-n^2
      Tn=n^-2 - (n+1)^-2
      Summation to n terms is
      Sn= 1-(n+1)^-2
      putting n=9 to the given problem
      We get S9=99/100

      2)Tn for this series is
      Tn=1/n - 1/(n+1)
      Sn=1-1/(n+1)
      X=Sn=n/(n+1)

      3) It can be seen that

      7^1=7
      7^2=49
      7^3=...3
      7^4=...1
      7^5=...7
      7^5 ends in 7.so also 7^5*7^5=7^25.
      In general 7^5^n ends in 7.for given prob n=6^13.

      --- In ascent4cat@yahoogroups.com, "Tarsem Saini" <ts_chd@y...> wrote:
      > Hi All,
      > Please help me out to solve the following problems in CAT approach.
      >
      > 1)find the value of 3/4 + 5/36 + 7/144 + ..... + 17/5184 + 19 / 8100
      >
      > a) 0.96
      > b) 0.95
      > c) 0.99
      > d) None of These
      >
      > /******************************************************************/
      >
      > 2)Compute x if x = 1/1*2 + 1/2*3 + 1/3*4 + ....+ 1/(n-1)*n + 1/n*
      (n+1)
      >
      > a) n/(n-1)
      > b) 2n/(n+1)
      > c) n/(n+1)
      > d) None of these
      >
      > /******************************************************************/
      >
      > 3) What is the Unit digit of 7^5^6^13 ?
      >
      > I have been able to solve this question by calculating power and
      then
      > devide by 4 .But it take some time to calculate power. Please tell
      > the Shortcut for this..
      >
      > /******************************************************************/
      >
      > Take care and prepare well.
      >
      > Saini
    • vinu chandran
      hi 1 . the general form of the nth term is 2n + 1 = 2n + 1 = 1 [ 1 - n^2 ] = 1 - 1 (n(n+1))^2 n^2(n^2+2n+1) n^2
      Message 2 of 3 , Jul 30, 2003
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        hi
        1 . the general form of the nth term is
            2n + 1    2n + 1            = 1     [ 1 - n^2       ] = 1      -      1     
           (n(n+1))^2   n^2(n^2+2n+1)     n^2        (n+1)^2       n^2       (n+1)^2
        now as the value of n ranges from 1 to 9
        the given sum is  1 - 1    +  1     -  1     ......... 1      -      1  
                                              2^2     2^2     3^2            9^2       10^2
        =1 -   1     =0.99
               100
        so Ans - c)
         2. 1          =  1     -       1    
            n(n+1)       n          (n+1)
        here also as n ranges from 1 to n
        1 -  1 + 1  - .....  -    1    =  1 -    1      =   n    
              2    2              n+1            n+1        n+1
        so Ans - c)
         
        3. anyway 5^6^13 or even 5^25000 has same last 2 digits ie 25
           so the qn becomes smething like
           7^((n * 100) + 25) so this becomes 7^(4m +1)
           this gives a 7 in the units place
         
        bye
        g'luk everyone
         
         
        Tarsem Saini <ts_chd@...> wrote:
        Hi All,
        Please help me out to solve the following problems in CAT approach.

        1)find the value of 3/4 + 5/36 + 7/144 + ..... + 17/5184 + 19 / 8100

        a) 0.96
        b) 0.95
        c) 0.99
        d) None of These

        /******************************************************************/

        2)Compute x if x = 1/1*2 + 1/2*3 + 1/3*4 + ....+ 1/(n-1)*n + 1/n*(n+1)

        a) n/(n-1)
        b) 2n/(n+1)
        c) n/(n+1)
        d) None of these

        /******************************************************************/

        3) What is the Unit digit of 7^5^6^13 ?

        I have been able to solve this question by calculating power and then
        devide by 4 .But it take some time to calculate power. Please tell
        the Shortcut for this..

        /******************************************************************/

        Take care and prepare well.

        Saini






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