Question

如何求2^x=x^2的解?为什么1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6?

Answer 1

1^2 2^2 3^2 …… n^2=n(n 1)(2n 1)/6

利用立方差公式

n^3-(n-1)^3=1*[n^2 (n-1)^2 n(n-1)]

=n^2 (n-1)^2 n^2-n

=2*n^2 (n-1)^2-n

2^3-1^3=2*2^2 1^2-2

3^3-2^3=2*3^2 2^2-3

4^3-3^3=2*4^2 3^2-4

.

n^3-(n-1)^3=2*n^2 (n-1)^2-n

各等式全相加

n^3-1^3=2*(2^2 3^2 ...n^2) [1^2 2^2 ...(n-1)^2]-(2 3 4 ...n)

n^3-1=2*(1^2 2^2 3^2 ...n^2)-2 [1^2 2^2 ...(n-1)^2 n^2]-n^2-(2 3 4 ...n)

n^3-1=3*(1^2 2^2 3^2 ...n^2)-2-n^2-(1 2 3 ...n) 1

n^3-1=3(1^2 2^2 ...n^2)-1-n^2-n(n 1)/2

3(1^2 2^2 ...n^2)=n^3 n^2 n(n 1)/2=(n/2)(2n^2 2n n 1)

=(n/2)(n 1)(2n 1)

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