计算(1²+3²+...+99²)-(2²+4²+...+100&su
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/23 19:09:52
计算(1²+3²+...+99²)-(2²+4²+...+100²)急!
![计算(1²+3²+...+99²)-(2²+4²+...+100&su](/uploads/image/z/18442619-35-9.jpg?t=%E8%AE%A1%E7%AE%97%EF%BC%881%26sup2%3B%2B3%26sup2%3B%2B...%2B99%26sup2%3B%EF%BC%89-%EF%BC%882%26sup2%3B%2B4%26sup2%3B%2B...%2B100%26su)
本题应用“平方差公式”:a²-b²=(a+b)(a-b)
有了平方差公式,问题就好解决了:
(1²+3²+...+99²)-(2²+4²+...+100²)
=1²+(3²-2²)+(5²-4²)...+(99²-98²)-100²
=1-10000+(3+2)(3-2)+(5+4)(5-4)+...+(99+98)(99-98)
=1-10000+(3+2)+(5+4)+...+(99+98)
=1+2+3+4+...+98+99-10000
用“等差数列求和的方法”[(首项+莫项)*项数/2 ],得:
(1+99)*99/2-10000
=100*99/2-10000
=9900/2-10000
=4950-10000
= - 5050
有了平方差公式,问题就好解决了:
(1²+3²+...+99²)-(2²+4²+...+100²)
=1²+(3²-2²)+(5²-4²)...+(99²-98²)-100²
=1-10000+(3+2)(3-2)+(5+4)(5-4)+...+(99+98)(99-98)
=1-10000+(3+2)+(5+4)+...+(99+98)
=1+2+3+4+...+98+99-10000
用“等差数列求和的方法”[(首项+莫项)*项数/2 ],得:
(1+99)*99/2-10000
=100*99/2-10000
=9900/2-10000
=4950-10000
= - 5050
(2²+4²+6²+.+98²+100²)-(1²+3&su
计算:100²-99²+98²-97²+…4²-3²+2&s
计算:100²-99²+98²-97²+……+4²-3²+2
100²-99²+98²-97²+.4²-3²+2²
1²-2²+3²-4²+5²-6²+…-100²+
计算100²-99²-+98²-97²+...+2²-1
计算100²-99²+98²-97²+...+2²-1
100²-99²-98²-97²-.-1²
乘法公式计算100²-99²+98²-97²+……+2²-1²
1.计算:1²+4²+6²+7²=102,2²+3²+5&s
100²-99²+98²-97²+.+4²-3²-3&sup
100²-99²+98²-97²+96²+……+2²-1&s