女巫在身边图片:1×1+2×2+......+n×n=?
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关于 1^2 + 2^2 + …… + n^2 = n×(n+1)×(2n+1)/6 的证明:
(n+1)^3 = n^3 + 3n^2 + 3n + 1
所以
(n+1)^3 - n^3 = 3n^2 + 3n + 1
n^3 - (n-1)^3 = 3(n-1)^2 + 3(n-1) + 1
(n-1)^3 - (n-2)^3 = 3(n-2)^2 + 3(n-2) + 1
............
3^3 - 2^3 = 3*2^2 + 3*2 + 1
2^3 - 1^3 = 3*1^2 + 3*1 +1
把以上n个等式的两边分别相加得到
(n+1)^3-1^3 =
3×(1^2+2^2+3^2+...+n^2) + 3×(1+2+3+……+n) + n个1的和
(n+1)^3-1 = 3×(1^2+2^2+...+n^2) + 3×n×(n+1)/2 + n
所以
3(1^2+2^2+......+n^3)
= n^3 + 3n^2 + 3n - 3n(n+1)/2 - n
= n(n^2+3n+2) - 3n(n+1)/2
= n(n+1)(n+2)-3n(n+1)/2
= n(n+1)(2n+1)/2
最后
1^2+2^2+......+n^2 = n(n+1)(2n+1)/6.
n(n+1)(2n+1)/6
"1^n+2^n+3^n......+m^n=?
1^n+2^n+3^n......+m^n=
x=n*(n+1)*(n+2)*(n+3).......
n/(n-2)+n/(n-3)+n/(n-4)+n/(n-5)+...+2/(-1)=?
lim(n→∞)[1/(n^2+n+1)+2/(n^2+n+2)+.....+n/(n^2+n+n)]=?
(n+2)^2-n^2=4*(n+1)
证明C(0,n)^2+C(1,n)^2+……+C(n,n)^2=C(n,2n)
1*1!+2*2!+~~~~~~~~~+n*n!=?
1×1+2×2+......+n×n=?
若f(n)=(n+1)(n+2)(n+3)+……(n+n),求f(n+1)/f(n)