Mathematical Induction~

[匿名]

2010-11-13 8:21 am

a)Prove, by mathematical induction, that
1x4+2x7+3x10+...+n(3n+1)=n(n+1)^2
for all positive integers n.

b)Hence, evaluate 1x5+2x8+3x11+...+99x299.

最主要唔識b),Please help me!!

回答 (1)

用戶2053

2010-11-13 8:34 am

✔ 最佳答案

a)1x4+2x7+3x10+...+n(3n+1)=n(n+1)^2When n = 1 , 1 x 4 = 1(1+1)^2 is true.Assume that when n = k
i.e. 1x4+2x7+3x10+...+k(3k+1)=k(k+1)^2 the statement is true.When n = k+1 :1x4+2x7+3x10+...+k(3k+1) + (k+1)(3(k+1)+1)
= k(k+1)^2 + (k+1)(3(k+1)+1)= k(k+1)^2 + (k+1)(3k+4)= (k+1)(k(k+1) + 3k+4)= (k+1)(k^2 + 4k + 4)= (k+1)(k+2)^2 is true.By Mathematical Induction it is true for any integers.b)1x5+2x8+3x11+...+99x299= 1x(4+1) + 2x(7+1) + 3x(10+1) + ... + 99x(3*99+1 + 1)= (1x4 + 1) + (2x7 + 2) + (3x10 + 3) + ... + 99x(3*99+1) + 99= 1x4+2x7+3x10+...+99(3*99+1) + (1 + 2 + 3 + ... + 99)= 99(99+1)^2 + (1+99)99/2= 994950

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