amath 一問 數學歸納法

1x32+2x52+3x72+...+n(2n+1)2=n(n+1)(6n2+14n+7) /6 ...........
Therefore, by principle of mathematical induction, expression is true for all positive integer n
當n=1
LHS = 1*32 = 9
RHS = n(n+1)(6n2+14n+7) /6 = 1(1+1)(6*12+14*1+7)/6 = 9
故當n=1時成立
設 n = k 時成立
則
1x32+2x52+3x72+...+k(2k+1)2=k(k+1)(6k2+14k+7) /6
當n = k + 1
則
LHS = 1x32+2x52+3x72+...+k(2k+1)2+(k+1)(2k+3)2
= k(k+1)(6k2+14k+7) /6 + (k+1)(2k+3)2
= k(k+1)(6k2+14k+7) /6 + 6(k+1)(2k+3)2/6
= (k+1)[k(6k2+14k+7) + 6(2k+3)2]/6
= (k+1)[6k3+14k2+7k + 6(4k2+12k+8)]/6
= (k+1)[6k3+14k2+7k + 24k2+72k+48]/6
= (k+1)[6k3+38k2+79k+48]/6
= (k+1)(k+2)[6k2+26k+27]/6
RHS = (k+1)(K+2)(6(k+1)2+14(k+1)+7) /6
= (k+1)(K+2)(6(k2+2k+1)+14k+14+7) /6
= (k+1)(K+2)(6k2+26k+27) /6
LHS = RHS
所以對任何整數n均成立

hence,or otherwise,find the sum of
11x232+12x252+13x272+...+20x412
1x32 + 2x52 + ……....+20x412
= n(n+1)(6n2+14n+7) /6
= (20)[(20)+1][6(20)2+14(20)+7] /6
= (20)[21][6(400)+280+7] /6
= (20)[21][6(400)+280+7] /6
= 188090
1x32+2x52+3x72+……+10x212
= n(n+1)(6n2+14n+7) /6
= (10)[(10)+1][6(10)2+14(10)+7] /6
= (10)[11][6(100)+140+7] /6
= (10)[11][6(100)+140+7] /6
= 13695
所以
11x232+12x252+13x272+...+20x412
= 188090 – 13695
= 174395

我唔係好明點解
11x232+12x252+13x272+...+20x412=1x32+2x52+3x72+...+20x412-(1x32+2x52+3x72+........+10x212)


因為
1x32+2x52+3x72+…+10x212+11x232+12x252+13x272+...+20x412
將這數列分為兩部份，n由1至10及由11至20兩部份
所以
11x232+12x252+13x272+...+20x412 可以等於由
1x32+2x52+…. +20x412減去前半部1x32+2x52+3x72+…+10x212

計錯數

聽聞知識長計錯數，我來睇下．

1×3^2+2×5^2+3×7^2+...+n(2n+1) ^2=1/6×n(n+1)(6n^2+14n+7)
Let P(n) be the proposition of “1×3^2+2×5^2+3×7^2+...+n(2n+1) ^2=1/6×n(n+1)(6n^2+14n+7)”
When n=1 LHS=1(2×1+1)^2
=9
RHS=1(1+1)(6×1^2+14x1+7)/6
=2(6+14+7)/6
=9
=LHS
∴P(1) is true
For any positive integer k, assume that P(k) is true.
ie 1×3^2+2×5^2+3×7^2+...+k(2k+1)^ 2=k(k+1)(6k^2+14k+7)/6
Then 1×3^2+2×5^2+3×7^2+...+k(2k+1) ^2+(k+1)【2(k+1)+1】^2
= k(k+1)(6k^2+14k+7)/6+【2(k+1)+1】^2
=(k+1)【k(6k^2+14k+7)/6+(2k+3)^2】
=(k+1)(6k^3+14k^2+7k+24k^2+72k+54)/6
=(k+1)(6k^3+38k^2+79k+54)/6
=(k+1)(6k^3+26k^2+27k+12k^2+52k+54)/6
=(k+1)(k+2)(6k^2+26k+27)/6
=(k+1)(k+2)(6k^2+12k+6+14k+14+7)/6
=(k+1)(k+2)【6(k^2+2k+1)+14(k+1)+7】/6
=(k+1)(k+2)【6(k+1)^2+14(k+1)+7】/6
Thus for any positive integer k, assume that P(k) is true, then P(k+1) is true.
So by the principle of Mathematical Induction, P(n) is true for any positive interger.
Hence or otherwise, find 11×23^2+12×25^2+13×27^2+...+20x41^2
11×23^2+12×25^2+13×27^2+...+20x41^2
=(1x3^2+2x5^2+3x7^2+...+20x41^2)-(1x3^2+2x5^2+3x7^2+........+10x21^2)
=20(20+1)(6x20^2+14x20+7)/6-10(10+1)(6x10^2+14x10+7)/6
=1128540/6-82170/6
=188090-13695
=174395
因為(1x3^2+2x5^2+3x7^2+...+20x41^2)-(1x3^2+2x5^2+3x7^2+........+10x21^2)當中的(1x3^2+2x5^2+3x7^2+...+20x41^2) 代表了(1x3^2+2x5^2+3x7^2+........+10x21^2+........+20x41^2)
試想想當
(1x3^2+2x5^2+3x7^2+........+10x21^2+11x23^2+........+20x41^2)-(1x3^2+2x5^2+3x7^2+........+10x21^2)時，有甚麼會剩下？剩下的當然是
(11x23^2+12x25^2+13x27^2+...+20x41^2)
所以若(1x3^2+2x5^2+3x7^2+...+20x41^2)-(1x3^2+2x5^2+3x7^2+........+10x21^2)，結果便會是
(11x23^2+12x25^2+13x27^2+...+20x41^2)了。

2006-10-28 22:03:54 補充：
在中間的(k+1)(6k^3+38k^2+79k+54)/6用正常方法是無辦法繼續下去的，但是當你用已知的答案(k+1)(k+2)【6(k+1)^2+14(k+1)+7】/6從答案做上問題時，你便可輕易地從(k+1)(k+2)【6(k+1)^2+14(k+1)+7】/6變成(k+1)(6k^3+38k^2+79k+54)/6 了而我在中間的正是我由答案變成問題的步驟了當然，你在真正做數時，便一定要跟著我在正文的步驟做。

2006-10-28 22:05:25 補充：
在中間的(k 1)(6k^3 38k^2 79k 54)/6用正常方法是無辦法繼續下去的，但是當你用已知的答案(k 1)(k 2)【6(k 1)^2 14(k 1) 7】/6從答案做上問題時，你便可輕易地從(k 1)(k 2)【6(k 1)^2 14(k 1) 7】/6變成(k 1)(6k^3 38k^2 79k 54)/6 了而我在中間的正是我由答案變成問題的步驟了當然，你在真正做數時，便一定要跟著我在正文的步驟做。因為你在做數時，要假設自己是不知道答案的。