MATHEMATICAL INDUCTION

4)
for n = 1,
　(1)(1 + 1)(2 + 1)
= 6　　　which is divisible by 3
∴ the statement is ture for n = 1
Assume the statement is ture for n = k for some integer k ≥ 1
i.e. k(k + 1)(2k + 1) = 3M　for some integer M
Then, n = k + 1
　(k + 1)(k + 2)(2k + 3)
= 2k(k + 1)(k + 2) + 3(k + 1)(k + 2)
= 2(3M) + 3(k + 1)(k + 2)
= 3[2M + (k + 1)(k + 2)]　　　which is divisible by 3
∴ the statement is ture for n = k + 1
By the principle of mathematical induction,
the statment is true for all natural numbers n.

18)
3×5 + 7×9 + 11×13 + … + (4n - 1)(4n + 1) = 1/3×n(4n + 1)(4n + 5)
(i) 3×5 + 7×9 + 11×13 + … + 95×97
= 3×5 + 7×9 + 11×13 + … + [4(24) - 1] [4(24) + 1]
= (24)[4(24) + 1][4(24) + 5] /3
= 78376
(ii) 119×121 + 123×125 + … + 163×165
= 3×5 + 7×9 + … + 163×165 - (3×5 + 7×9 + … + 115×117)
= 3×5 + 7×9 + … + [4(41) - 1] [4(41) + 1] - {3×5 + 7×9 + … + [4(29) - 1] [4(29) + 1]}
= (41)[4(41) + 1][4(41) + 5] /3 - (29)[4(29) + 1][4(29) + 5] /3
= 381095 - 136851
= 244244

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)

21)
是否打錯題目
根本計唔到

2007-01-12 18:46:05 補充：
上面的4) 並非證n(n 1)(n 2) is divisible by 3而係證n(n 1)(2n 1) is divisible by 321) 6^n(5n-1) 1題目何來有n次

2007-01-13 09:15:52 補充：
第4題有少少野錯左下面先岩　(k 1)(k 2)(2k 3)= (k 1)(k 2)(2k 1 2)= (k 1)(k 2)(2k 1) 2(k 1)(k 2)= k(k 1)(2k 1) 2(k 1)(2k 1) 2(k 1)(k 2)= 3M 2(k 1)[(2k 1) (k 2)]= 3M 2(k 1)(3k 3)= 3M 6(k 1)^2= 3[M 2(k 1)^2]　　　which is divisible by 3

2007-01-13 09:17:24 補充：
下面的人兄你20(a)錯了這步=1(5) 2(6) 3(7) ... n(n 4)-4(1 2 3 ...n)=1/6(n)(n 1)(2n 13)-1/2(n)(n 1)　　　　　　　　　↑應該要乘四的