(1^4+2^4+……+99^4+100^4)/(1^2+2^2+……+99^2+100^2)=?
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/04/28 00:12:23
(1^4+2^4+……+99^4+100^4)/(1^2+2^2+……+99^2+100^2)=?
“^”是次方,“/”是除号
“^”是次方,“/”是除号
1²+2²+3³+…+n²的和
∵(n+1)³-n³=3n²+3n+1
∴n=1时有 2³-1³=3*1²+3*1+1
n=2时有 3³-2³=3*2²+3*2+1
n=3时有 4³-3³=3*3² +3*3+1
.
n=n时有 (n+1)³-n³=3*n² +3*n+1
将以上n个等式竖向相加,得:
(n+1)³-1=3(1²+2²+3²+...+n²)+3(1+2+3+...+n)+n
=3(1²+2²+3²+...+n²)+3(n+1)n/2+n
=3(1²+2²+3²+...+n²)+n(3n+5)/2
∴ 1²+2²+3²+...+n²=(1/3)[(n+1)³-n(3n+5)/2-1]
=(1/3)[n³+3n²+3n-(3n²+5n)/2]
=(1/6)(2n³+3n²+n)
=n(2n²+3n+1)/6
=n(n+1)(2n+1)/6
所以当n=100时 1²+2²+3²+…+100²=100(100+1)(200+1)/6
当n=100²时(1²)²+(2²)²+(3²)²+…+(100²)=100²(100²+1)(2×100²+1)/6
所以((1²)²+(2²)²+(3²)²+…+(100²))/(1²+2²+3²+…+100²)=[100²(100²+1)(2×100²+1)/6]/[100(100+1)(200+1)/6]
∵(n+1)³-n³=3n²+3n+1
∴n=1时有 2³-1³=3*1²+3*1+1
n=2时有 3³-2³=3*2²+3*2+1
n=3时有 4³-3³=3*3² +3*3+1
.
n=n时有 (n+1)³-n³=3*n² +3*n+1
将以上n个等式竖向相加,得:
(n+1)³-1=3(1²+2²+3²+...+n²)+3(1+2+3+...+n)+n
=3(1²+2²+3²+...+n²)+3(n+1)n/2+n
=3(1²+2²+3²+...+n²)+n(3n+5)/2
∴ 1²+2²+3²+...+n²=(1/3)[(n+1)³-n(3n+5)/2-1]
=(1/3)[n³+3n²+3n-(3n²+5n)/2]
=(1/6)(2n³+3n²+n)
=n(2n²+3n+1)/6
=n(n+1)(2n+1)/6
所以当n=100时 1²+2²+3²+…+100²=100(100+1)(200+1)/6
当n=100²时(1²)²+(2²)²+(3²)²+…+(100²)=100²(100²+1)(2×100²+1)/6
所以((1²)²+(2²)²+(3²)²+…+(100²))/(1²+2²+3²+…+100²)=[100²(100²+1)(2×100²+1)/6]/[100(100+1)(200+1)/6]
(1^4+2^4+……+99^4+100^4)/(1^2+2^2+……+99^2+100^2)=?
1 2 3 4…… 99 100=?
(-1)+(+2)+(-3)+(+4)+(-5)+……+(-99)+(+100)=?
(+1)+(-2)+(+3)+(-4)+……+(+99)+(-100)=_____
1+2-3+4-……+99-100=?
(-1)+2+(-3)+4+…+(-99)+100
(100+98+96+……+4+2)-(99+97+95+……+3+1)
1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4)……+1/(1+2+3+4+……+99+100)=(
(-9)+(-19)+(-29)+……+(99) 和(+1)+(-2)+(+3)+(-4)+……+(+99)+(-100
1+2+3+4……+98+99+100=()x()
1+(-2)+3+(-4)+5+(-6)+……+(99)+(-100)=?
计算:(+1)+(-2)+(+3)+(-4)+……+(+99)+(-100)+(+101).