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Today’s Math Question                                        2009   11   8
        0002
                             a                      n          (n + 1)
                                    b
r

(1) b a      n   r
(2) r = 0.08          b 2a                                 n
                     log10 2 = 0.3010   log10 3 = 0.4771
Today’s Math Answer
            0002
(1)                  a                               a(1 + r)       n   (2)                        b     2a

                                                                                             a{(1 + r)n − 1}       2a
       a(1 + r)n−1
                                                                                                          a > 0                   a
                            n                                                 r = 0.08
      S
                                                                                                 (1 + r)n − 1      2
                   n−1              n−2
      S = a(1 + r)        + a(1 + r)       + ...                                                         (1.08)n   3
                                                 1         0
                                 + a(1 + r) + a(1 + r)

           a         (1 + r)           n
                                                                                                n log10 (1.08)     log10 3
                                                                                 n{log10 (33 × 22 × 10−2 )}        log10 3
         a{(1 + r) − 1}
                      n
                          a{(1 + r) − 1}     n
      S=                =                                                       n(3 log10 3 + 2 log10 2 − 2)       log10 3
           (1 + r) − 1           r
                                                                                                                              0.4771
                     (n + 1)                                    b                                             n
                                                                                                                   3 × 0.4771 + 2 × 0.3010 − 2
                                                                                                              n    14.3 . . .
      b = Sr = a{(1 + r)n − 1} . . . . . .
                                                                                         n   15 . . . . . .

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0002

  • 1. Today’s Math Question 2009 11 8 0002 a n (n + 1) b r (1) b a n r (2) r = 0.08 b 2a n log10 2 = 0.3010 log10 3 = 0.4771
  • 2. Today’s Math Answer 0002 (1) a a(1 + r) n (2) b 2a a{(1 + r)n − 1} 2a a(1 + r)n−1 a > 0 a n r = 0.08 S (1 + r)n − 1 2 n−1 n−2 S = a(1 + r) + a(1 + r) + ... (1.08)n 3 1 0 + a(1 + r) + a(1 + r) a (1 + r) n n log10 (1.08) log10 3 n{log10 (33 × 22 × 10−2 )} log10 3 a{(1 + r) − 1} n a{(1 + r) − 1} n S= = n(3 log10 3 + 2 log10 2 − 2) log10 3 (1 + r) − 1 r 0.4771 (n + 1) b n 3 × 0.4771 + 2 × 0.3010 − 2 n 14.3 . . . b = Sr = a{(1 + r)n − 1} . . . . . . n 15 . . . . . .