1^2 + 2^2 + 3^2 + ... + n^2 點解 = 1/6 n (n+1) (n+2)

2007-09-11 7:57 am
1^2 + 2^2 + 3^2 + ... + n^2 點解 = 1/6 n (n+1) (n+2)
要step

回答 (3)

2007-09-11 8:24 am
✔ 最佳答案
lee類數可以用Mathematical Induction去計
當n=1,
LHS=1^2=1
RHS=1/6*(1)*(1+1)*(1+2)=1
所以當n=1以上句子是對的

我們假設當n=k時以上句子是對的
即:1^2+2^2+...+k^2=1/6*k*(k+1)*(k+2)
當n=k+1
LHS=1^2+2^2+...+k^2+(k+1)^2
=1/6*k*(k+1)*(k+2)+(k+1)^2
=1/6*(k+1)*[k*(k+2)+6*(k+1)]
=1/6*(k+1)*(k^2+8k+6)
=1/6*(k+1)*(k+1+1)*(k+2+1)
=RHS
所以當n=k+1以上句子都是對的

By the principle of mathematical inductinon, the proposition is true for all positive integer n

希望幫到你
參考: own
2007-09-21 9:10 am
問題係問點解,無叫你去prove.
解左既無人投最佳,無解既反而攞最佳.
原來呢度係攪笑+ 唔係知識+
2007-09-11 3:22 pm
first

Assume 1^2 + 2^2 + 3^2 + ... + n^2 = Q

(k+1)^3-k^3=3k^2+3k+1

2^3-1^3=3X1^2+3X1+1
3^3-2^3=3X2^2+3X2+1
4^3-3^3=3X3^2+3X3+1
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n^3-(n-1)^3=3(n-1)^2+3(n-1)+1
(n+1)^3-n^3=3n^2 +3n +1

add all of them together

(n+1)^3-1^3=3Q+3[n(n+1)/2]+n

Then 3Q=(n+1)^3-3[n(n+1)/2]-n-1
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Q=n(n+1)(2n+1)/6
參考: A stupid guy from Macau


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