For any positive integer n, the sum of the first n positive integers equals ~$\frac{n\left ( n+1 \right )}{2}$~. What is the sum of all the even integers between 99 and 301?
sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50
Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101
The sum = 200*101= 20,200.
1~99的和为99*(99+1)/2=4950
1~301的和为301*(301+1)/2=45451
100~301的和为45451-4950=40501
100~301 设偶数和为x 则奇数和为x+101 (共202个数,奇数和偶数各101个 奇数和比偶数和大101)
x+(x+101)=40501
x=20200
即偶数和为20200
等于求A1=100, An=300, D=2 的等差数列和。 因为N=(300-100)除以2 然后加上1, 所以N =101, Sn= (100+300)*101/2= 200*101= 20200
sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50
required sum = 150*151 - 49*50 = 50*(453 - 49) =
404 * 50 = 20200
between and 包含两边
项数用通项公式算,求和用求和公式算,连续偶数和是等差数列
注意:偶数的个数是(300-100)÷2+1!!!
the sum of all the even integers 求偶数和,要÷2!!!
第一句话没有用。用等差数列求和公式即可。
奇数偶数和,先看总个数
1、总个数为n偶数,最后一项是奇数,则奇数数和大于偶数和n/2;最后一项是偶数,偶数和大于奇数和n/2;
2、总个数为n奇数,最后一项是奇数,则奇数数和大于偶数和(个数多的中位数);最后一项是偶数,偶数和大于奇数和(个数多的中位数);
太清晰了!谢谢
1
登录 或 注册 后可以参加讨论
1~99的和为99*(99+1)/2=4950
1~301的和为301*(301+1)/2=45451
100~301的和为45451-4950=40501
100~301 设偶数和为x 则奇数和为x+101 (共202个数,奇数和偶数各101个 奇数和比偶数和大101)
x+(x+101)=40501
x=20200
即偶数和为20200
https://www.zybang.com/question/12d3b445fc745ce5da0050fa5ac94226.html
the sum of the first n positive integers 前n个正整数的和