前50个偶数的和是多少?
算术是数学的一部分,它处理不同类型的数字、分数,对数字应用不同的运算,如加法、乘法等。算术这个词来自希腊词 arithmos,意思是数字。算术还包括求幂、百分比计算、求数列的值、对数函数和平方根等。
算术中有一个称为算术级数(AP)的序列,这是一个数字序列,其中任何两个连续项之间的差异总是相同的。比方说,一个数列是 2,4,6,8,10,12,…..,在这个数列中,任何两个连续数字之间的差是 2。如果我们将这个 2 与前一个数字相加,那么我们得到下一个序列中的数字,类似地,如果我们从下一个数字中减去 2,我们会得到前一个数字。
在这篇文章中,我们找到了前 50 个偶数的总和。要使用这个系列,有一些公式可用,即,
假设一个系列 A 由一些元素组成 a 1 , a 2 , a 3 , a 4 , a 5 , a 6 ,...a n
A = {a 1 , a 2 , a 3 , a 4 , a 5 , a 6 ,…a n }
- 两项的共同点 (d) = (a 1 -a 2 )
- 系列之和 (S) = (n/2)[2a + (n – 1)d]
- 第一项 = 一个
- 第二项 = a+d
- 第三项 = a+2d
- 同样,第 N 项 = a+(n-1)d
现在让我们取一个有 5 个项的系列,那么系列的总和如下所示,
1st term, a1 = a
2nd term, a2 = a+d
3rd term, a3 = a+2d
4th term, a4 = a+3d
5th term, a5 = a+4d
so, sum of this series, S5= (a1 +a2 +a3 +a4+a5) = (a + a+d + a+2d + a+3d + a+4d)
It can also be written as,
=> S5 = a+a+a+a+a + d+2d+3d+4d
=> S5 = 5a+10d
=> S5 = 5×(a + 2d) [take 5 as common] ——————-(1)
we can this equation as,
2×S5 = 5 × 2 × [a + 2d] [multiply 2 in both side]
=> 2×S5 = 5×[2a+4d]
=> S5 = (5/2)×[2a + (5-1)d] Here n = 5,
现在让我们取一个有 4 项的系列,那么系列的总和如下所示,
1st term, a1 = a
2nd term, a2 = a+d
3rd term, a3 = a+2d
4th term, a4 = a+3d
so, sum of this series, S4 = (a + a+d + a+2d + a+3d)
It can also be written as,
=> S4 = a+a+a+a + d+2d+3d
=> S4 = 4a+6d
=> S4 = 2×(2a + 3d) [take 2 as common]
we can this equation as,
S4 = (4/2) × [2a + (4-1)*d] Here n = 4,
因此,为了找到具有 N 项的系列之和,公式如下所示
Sn = (n/2)×[2a + (n-1)×d]
前50个偶数的和是多少?
First of all, we have to find the series of even numbers, let’s find that
The first even number is 2,
2nd even number is 4
3rd even number is 6 and so on….. so the last term will be 50 × 2 = 100
Secondly, find the common difference between them, d = 4 – 2 = 2 or 6 – 4 = 2
So, d = 2,
Thirdly, find the sum of the first 50 even terms,
Use the formula for finding the sum of series, that is Sn = (n/2)×[2a + (n-1)×d]
here, n=50, a=2, d=2
therefore, S50 = (50/2)×[2×2 + (50-1)×2]
=> S50 = 25×[4+ 49×2]
=> S50 = 25×102
=> S50 = 2550
The sum of first 50 even number is 2550
另一种解决方法:
To find the sum of even numbers up to N we can use another formula, that is, N×(N+1)
This formula only works with consecutive even numbers, that is, 2,4,6,8,10,12,14,16…… like this.
Let’s say a series has the first 5 even numbers, 2,4,6,8,10
so the sum of this series,
S5 = 5×(5+1) [N =5]
S5 = 5×6 = 30
To check if the sum is correct or not we can sum up the numbers, 2+4+6+8+10 = 30, so the formula works fine
Here we are finding the sum of first 50 even numbers,
The sum, S50 = N*(N+1)
S50 = 50*(50+1) = 50*51
S50 = 2550
The sum of first 50 even number is 2550
另一种解决方法:
For this method, we need to find the last term of the series.
To find the last term we a formula,
Tn = a+(n-1)d
=> T50 = 2+(50-1)×2 [Here, a=2, d=2. n=50]
=> T50 = 2+49×2 = 2+98
=> T50 = 100 ———–(2)
Now to find the sum, the formula is, Sn=(N/2) × (a + Tn), This formula is correct for all series,
let’s say a series has 4 terms 1,5,9,13. the sum of this series according to this formula will be,
S4 = (4/2)×(1+13) = 2 × 14 = 28
Check if it is correct or not, 1+5+9+13 = 28, so the formula is correct.
Now find the sum of the first 50 even numbers,
S50 = (50/2) × (2 + 100) [Here, a=2, Tn=100 — from (2)]
S50 = 25 × 102
S50 = 2550
The sum of first 50 even number is 2550
类似问题
问题 1. 求前 30 个偶数之和。
解决方案:
To find this sum we can use any previously defined method,
Let’s use the 2nd method, which is N×(N+1)
S30 = 30×(30+1)
=> S30 = 30×31
=>S30 = 930
The sum of the first 30 even numbers is 930
问题 2. 求一个数列的和,其第一项为 6,公差为 4,该数列的项数也为 6。
解决方案:
To solve this problem we can use the 3rd and 1st methods.
Using 1st method:
Given, d=4,a=6,N=6
put all the values in this formula Sn = (n/2)×[2a + (n-1)×d]
S6 = (6/2)×[2×6+(6-1)×4] = 3×(12+20)
S6 = 3×32
S6 = 96
Using 3rd method:
Given, d=4,a=6,N=6
Find the nth term,
Tn=a+(n-1)d
T6 = 6 + (6-1)×4
T6 = 6+ 20
T6 = 26——–nth term
Now put all the values in this formula Sn=(N/2) × (a + Tn)
S6 = (6/2) × (6+26)
S6 = 3×32
S6 = 96
Now find the series, first term is 6 and common difference is 4
so the series will be,
a1 = 6, a2 = 6+4 = 10, a3 = 10+4 =14,
a4 = 14+4 = 18, a5= 18+4= 22, a6= 22+4 = 26
the series is 6,10,14,18, 22, 26
the sum is 6+10+14+18+22+26 = 96
So the solution is correct. and the sum of this series is 96
问题 3. 求和为 147、最后一项为 33、该系列项数为 7 的数列。
解决方案:
Given, Sn =147, Tn = 33, N=7
we can use this Sn=(N/2) × (a + Tn) formula here to find the first term in that series
Putting all the given values, we get
147 = (7/2)×(a + 33)
=> 147×2 = 7×a + 7×33
=> 294 = 7a + 231
=> 7a = 294-231
=>7a = 63
=> a = 63/7
=>a = 9
So the first term is 9.
Now find the common difference d, to find d we can use any formula which contains d
Here we are using the formula for finding the Nth term, that is Tn=a+(n-1)d
Putting all the values, we get,
33 = 9 + (7-1)×d [Given Tn = 33, a=9, n=7]
=> 33-9 = 6×d
=>6d = 24
=>d=24/6
=> d=4
the common difference is 4
Now find the series,
a1 = 9, a2 = 9+4 = 13, a3 = 13+4 = 17, a4 = 17+4= 21, a5 = 21+4=25, a6 = 25+4 = 29, a7 = 29+4 = 33
So the final series is 9, 13, 17, 21, 25, 29, 33