MATHEMATICAL INDUCTION

20)
1×5 + 2×6 + 3×7 + … + n(n + 4) = 1/6 n(n + 1)(2n + 13)
and 1 + 2 + 3+ … + n = 1/2 n(n + 1)
(a) 1² + 2² + 3² + … + n²
= 1² + 2² + 3² + … + n² + 4(1 + 2 + 3 + … + n) - 4(1 + 2 + 3 + … + n)
= [(1² + 4) + (2² + 2×4) + (3² + 3×4) + … + (n² + 4n)] - 4[1/2 n(n + 1)]
= [1×5 + 2×6 + 3×7 + … + n(n + 4)] - 2n(n + 1)
= 1/6 n(n + 1)(2n + 13) - 2n(n + 1)
= 1/6 n(n + 1)[(2n + 13) - 12]
= 1/6 n(n + 1)(2n + 1)

22)
5^(2n - 1) - 3^(2n - 1) - 2^(2n - 1)
for n = 1,
　5^(2 - 1) - 3^(2 - 1) - 2^(2 - 1)
= 5 - 3 - 2
= 0　　　which is divisible by 15
∴ the statement is ture for n = 1
Assume the statement is ture for n = k for some integer k ≥ 1
i.e. 5^(2k - 1) - 3^(2k - 1) - 2^(2k - 1) = 15M　for some integer M
　　　　　　　　　　　 5^(2k - 1) = 15M + 3^(2k - 1) + 2^(2k - 1)
Then, n = k + 1
　5^(2k + 1) - 3^(2k + 1) - 2^(2k + 1)
= 5^2 × 5^(2k - 1) - 3^2 × 3^(2k - 1) - 2^2 × 2^(2k - 1)
= 25[5^(2k - 1)] - 9[3^(2k - 1)] - 4[2^(2k - 1)]
= 4[5^(2k - 1) - 3^(2k - 1) - 2^(2k - 1)] + 21[5^(2k - 1)] - 5[3^(2k - 1)]
= 4(15M) + 21[5^(2k - 1)] - 5[3^(2k - 1)]
= 60M + 21 × 5[5^(2k - 2)] - 5 × 3[3^(2k - 2)]
= 60M + 105[5^(2k - 2)] - 15[3^(2k - 2)]
= 15{4M + 7[5^(2k - 2)] - 3^(2k - 2)}　　　which is divisible 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.

2007-01-13 09:09:42 補充：
上面的22題人地叫你用MI啦

(a) n^2 = n(n)
= n(n+4) - 4n
= 1/6(n)(n+1)(2n+13) - 4[1/2(n)(n+1)]
= 1/6(n)(n+1)(2n+13) - 2(n)(n+1)
= [1/6(2n+13)-2] (n)(n+1)
= (1/3n+1/6)(n)(n+1)

(b) 5^2n-1 - 3^2n-1 - 2^2n-1
= 5^2n /5 - 3^2n /3 -2^2n /2
= 1/30 [ 6(5^2n) - 10 (3^2n) - 15 (2^2n) ]
= 15 (1/450) [ 6(5^2n) - 10 (3^2n) - 15 (2^2n) ]