Question

2^2-1=2×1+1，3^2-2^2=2×2+1,4^2-3^2=2×3+1......（n+1）-n^2=2n+1

Answer 1

（1）
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
各等式全部相加
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
（2）
由（1）知1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
∴2^2+4^2+...+(2n)^2
=2^2(1^2+2^2+..+n^2)
=
4n(n+1)(2n+1)/6
1^2+2^2+...+(2n+1)^2
=
(2n+1)(2n+2)(4n+3)/6
1^2+3^2+...+（2n+1）^2
=
1^2+2^2+...+(2n+1)^2
-
(
2^2+4^2+...+(2n)^2
)
=(2n+1)(2n+2)(4n+3)/6
-
4n(n+1)(2n+1)/6
=(2n+1)(
n+1)
(4n+3)/3
-
4n(n+1)(2n+1)/6
=[(n+1)(2n+1)/3
]
(
(4n+3)
-
2n
)
=
(n+1)(2n+1)
(2n+3)/3
1^2+3^2+5^2+...+99^2=50*99*101/3=50*33*101=166650.

Answer 2

证明：1+1/2^2+1/3^2+...+1/n^2<(2n-1)/n
(n>=2,n属于n*)
1）1+1/2^2=5/4
<
3/2
2)
设：1+1/2^2+1/3^2+...+1/k^2<(2k-1)/k,
1+1/2^2+1/3^2+...+1/k^2+1/(k+1)^2<(2k-1)/k+1/(k+1)^2
=(2k^3+4k^2+2k-k^2-2k-1+k)/k(k+1)^2
=2-(k-1)/k(k+1)<[2(k+1)-1]/(k+1)
也就是如果n=k时成立能推出n=k+1也成立
所以，1+1/2^2+1/3^2+...+1/n^2<(2n-1)/n