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0002
- 1. Today’s Math Question 2009 11 8
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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 . . . . . .