f.4 a maths -------------mathematical induction

For n = 1,
　5^(2 - 1) - 3^(2 - 1) - 2^(2 - 1)
= 5 - 3 - 2
= 0 which is divisble by 15
∴the statment is true for n = 1

Assume the statment is true for n = k for some positive integer k
i.e. 5^(2k - 1) - 3^(2k - 1) - 2^(2k - 1) = 15M for some integer M

When n = k + 1,
　5^[2(k + 1) - 1] - 3^[2(k + 1) - 1] - 2^[2(k + 1) - 1]
= 5^(2k + 1) - 3^(2k + 1) - 2^(2k + 1)
= 5^2 x 5^(2k - 1) - 3^2 x 3^(2k - 1) - 2^2 x 2^(2k - 1)
= 25 x 5^(2k - 1) - 9 x 3^(2k - 1) - 4 x 2^(2k - 1)
= 21[5^(2k - 1)] - 5[3^(2k - 1)] + 4[5^(2k - 1) - 3^(2k - 1) - 2^(2k - 1)]
= 21[5 x 5^(2k - 2)] - 5[3 x 3^(2k - 2)] + 4(15M)
= 105[5^(2k - 2)] - 15[3^(2k - 2)] + 60M
= 15{7[5^(2k - 2)] - 3^(2k - 2) + 4M} which is divisble by 15
∴the statment is true for n = k + 1
By the principle of mathematical induction, the statement is true for all positive integers n

2007-01-10 03:20:22 補充：
下面位人兄你唔使懶到直接copy呀自己打下字都得既

For n = 1,
　5^(2 - 1) - 3^(2 - 1) - 2^(2 - 1)
= 5 - 3 - 2
= 0 which is divisble by 15
∴the statment is true for n = 1

Assume the statment is true for n = k for some positive integer k
i.e. 5^(2k - 1) - 3^(2k - 1) - 2^(2k - 1) = 15M for some integer M

When n = k + 1,
　5^[2(k + 1) - 1] - 3^[2(k + 1) - 1] - 2^[2(k + 1) - 1]
= 5^(2k + 1) - 3^(2k + 1) - 2^(2k + 1)
= 5^2 x 5^(2k - 1) - 3^2 x 3^(2k - 1) - 2^2 x 2^(2k - 1)
= 25 x 5^(2k - 1) - 9 x 3^(2k - 1) - 4 x 2^(2k - 1)
= 21[5^(2k - 1)] - 5[3^(2k - 1)] + 4[5^(2k - 1) - 3^(2k - 1) - 2^(2k - 1)]
= 21[5 x 5^(2k - 2)] - 5[3 x 3^(2k - 2)] + 4(15M)
= 105[5^(2k - 2)] - 15[3^(2k - 2)] + 60M
= 15{7[5^(2k - 2)] - 3^(2k - 2) + 4M} which is divisble by 15
∴the statment is true for n = k + 1
By the principle of mathematical induction, the statement is true for all positive integers n