Đã cho một số n in tổng của tất cả các số chẵn từ 1 đến n trong C++
Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).5 Sum of first n even numbers = n * (n + 1).6 Sum of first 20 Even numbers is: 4204 Sum of first n even numbers = n * (n + 1).8 Show Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)0
Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)2Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)4Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)8Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4200 Sum of first 20 Even numbers is: 4201 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first 20 Even numbers is: 4203 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4205 Sum of first 20 Even numbers is: 4208 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4201____02 Sum of first 20 Even numbers is: 4203 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4205 Sum of first 20 Even numbers is: 4206 Sum of first 20 Even numbers is: 4207 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4200 Sum of first n even numbers = n * (n + 1).610 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6JavaSum of first n even numbers = n * (n + 1).612 Sum of first n even numbers = n * (n + 1).613 Sum of first n even numbers = n * (n + 1).614 Sum of first n even numbers = n * (n + 1).615 Sum of first n even numbers = n * (n + 1).614 Sum of first n even numbers = n * (n + 1).617
Sum of first n even numbers = n * (n + 1).618 Sum of first n even numbers = n * (n + 1).619 Sum of first n even numbers = n * (n + 1).700 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4202 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).70 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).707 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4205 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4207 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4208 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 4204 Sum of first n even numbers = n * (n + 1).736 Sum of first n even numbers = n * (n + 1).737 Sum of first n even numbers = n * (n + 1).738 Sum of first n even numbers = n * (n + 1).739 Sum of first n even numbers = n * (n + 1).770
Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).3 Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).5 Sum of first n even numbers = n * (n + 1).6 Sum of first 20 Even numbers is: 4204 Sum of first n even numbers = n * (n + 1).777 Sum of first n even numbers = n * (n + 1).778 Sum of first n even numbers = n * (n + 1).779 Sum of first 20 Even numbers is: 42000 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)0
Sum of first 20 Even numbers is: 42000 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)2Sum of first 20 Even numbers is: 42000 Sum of first 20 Even numbers is: 42005 Sum of first n even numbers = n * (n + 1).737 Sum of first n even numbers = n * (n + 1).770 Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)8Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 4200 Sum of first 20 Even numbers is: 42014 Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42019 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).618 Sum of first n even numbers = n * (n + 1).707 Sum of first 20 Even numbers is: 42023 Sum of first 20 Even numbers is: 42024 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4208 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 42029 Sum of first 20 Even numbers is: 42030 Sum of first n even numbers = n * (n + 1).770 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 42033____02 Sum of first 20 Even numbers is: 42035 ________ 036 ________ 06 ________ 038 Sum of first 20 Even numbers is: 42036 Sum of first 20 Even numbers is: 42040 Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first 20 Even numbers is: 42044 Python3Sum of first 20 Even numbers is: 42045 Sum of first 20 Even numbers is: 42046 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42048 Sum of first 20 Even numbers is: 42046 Sum of first 20 Even numbers is: 42050 Sum of first 20 Even numbers is: 42051 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42053 Sum of first 20 Even numbers is: 42054 Sum of first n even numbers = n * (n + 1).737 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42057 Sum of first 20 Even numbers is: 42054 Sum of first n even numbers = n * (n + 1).739 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42061 Sum of first 20 Even numbers is: 42054 Sum of first n even numbers = n * (n + 1).778 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42066 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42068 Sum of first 20 Even numbers is: 42069______054 Sum of first 20 Even numbers is: 42071 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 42057 Sum of first 20 Even numbers is: 42038 Sum of first 20 Even numbers is: 42054 Sum of first 20 Even numbers is: 42053 Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 42079 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 42053____038 Sum of first 20 Even numbers is: 42054 Sum of first n even numbers = n * (n + 1).737 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 42061 Sum of first 20 Even numbers is: 42054 Sum of first 20 Even numbers is: 42061 Sum of first 20 Even numbers is: 42038 Sum of first n even numbers = n * (n + 1).778 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4200 Sum of first 20 Even numbers is: 42057
Sum of first 20 Even numbers is: 42094 Sum of first 20 Even numbers is: 42095 Sum of first 20 Even numbers is: 42054 Sum of first 20 Even numbers is: 42030 Sum of first 20 Even numbers is: 42098 Sum of first n even numbers = n * (n + 1).6 Sum of first n even numbers = n * (n + 1).00 Sum of first n even numbers = n * (n + 1).01 Sum of first n even numbers = n * (n + 1).02 Sum of first n even numbers = n * (n + 1).03 Sum of first n even numbers = n * (n + 1).04 Sum of first n even numbers = n * (n + 1).05
Sum of first n even numbers = n * (n + 1).06 C#Sum of first n even numbers = n * (n + 1).07 Sum of first n even numbers = n * (n + 1).613 Sum of first n even numbers = n * (n + 1).77 Sum of first n even numbers = n * (n + 1).10
Sum of first n even numbers = n * (n + 1).618 Sum of first n even numbers = n * (n + 1).619 Sum of first n even numbers = n * (n + 1).13
Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4202 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).70 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).707 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4205 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4207 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4208 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 4204 Sum of first n even numbers = n * (n + 1).1
Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).3 Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).5 Sum of first n even numbers = n * (n + 1).6 Sum of first 20 Even numbers is: 4204 Sum of first n even numbers = n * (n + 1).8 Sum of first 20 Even numbers is: 42000 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)0
Sum of first 20 Even numbers is: 42000 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)2Sum of first 20 Even numbers is: 42000 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)4Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)8Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 4200 Sum of first 20 Even numbers is: 4201 Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 42019 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).618 Sum of first n even numbers = n * (n + 1).707 Sum of first 20 Even numbers is: 42023 Sum of first n even numbers = n * (n + 1).57 Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4208 Sum of first n even numbers = n * (n + 1).9 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).65 Sum of first 20 Even numbers is: 4202 Sum of first n even numbers = n * (n + 1).67 Sum of first 20 Even numbers is: 4204 Sum of first 20 Even numbers is: 42038 Sum of first 20 Even numbers is: 4206 Sum of first n even numbers = n * (n + 1).71 Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first n even numbers = n * (n + 1).75 PHPSum of first n even numbers = n * (n + 1).76 Sum of first n even numbers = n * (n + 1).77 Sum of first n even numbers = n * (n + 1).70
Sum of first 20 Even numbers is: 4202 Sum of first n even numbers = n * (n + 1).70 Sum of first n even numbers = n * (n + 1).81 Sum of first 20 Even numbers is: 4205____183 Sum of first n even numbers = n * (n + 1).84 Sum of first 20 Even numbers is: 4208 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).87 Sum of first n even numbers = n * (n + 1).88 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).90 Sum of first n even numbers = n * (n + 1).91
Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).3 Sum of first 20 Even numbers is: 4209 Sum of first n even numbers = n * (n + 1).5 Sum of first n even numbers = n * (n + 1).6 Sum of first n even numbers = n * (n + 1).97 Sum of first n even numbers = n * (n + 1).98 Sum of first n even numbers = n * (n + 1).97 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)00Sum of first n even numbers = n * (n + 1).83 Sum of first n even numbers = n * (n + 1).770 Sum of first n even numbers = n * (n + 1).97 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)04Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).90 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)07Sum of first n even numbers = n * (n + 1).87 Sum of first n even numbers = n * (n + 1).770
Sum of first n even numbers = n * (n + 1).9 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)2Sum of first n even numbers = n * (n + 1).9 Sum of first n even numbers = n * (n + 1).87 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)14Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)6
Sum of first 20 Even numbers is: 4209 Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.
Here, a = 2, d = 2, applying these values to eq.(i), we get
Sum = (n/2) * [2*2 + (n-1)*2]
= (n/2) * [4 + 2*n - 2]
= (n/2) * (2*n + 2)
= n * (n + 1)8Sum of first 20 Even numbers is: 4209 Sum of first 20 Even numbers is: 4200 Sum of first n even numbers = n * (n + 1).6 Sum of first 20 Even numbers is: 4204 Sum of first n even numbers = n * (n + 1).7311 Sum of first n even numbers = n * (n + 1).7312______2737 Sum of first n even numbers = n * (n + 1).84 Sum of first 20 Even numbers is: 42001 Sum of first n even numbers = n * (n + 1).6 Sum of first n even numbers = n * (n + 1).737 Sum of first 20 Even numbers is: 42038 Sum of first n even numbers = n * (n + 1).7319 Làm cách nào để in tổng các số chẵn trong C?Chương trình. Viết chương trình tìm tổng các số chẵn bằng ngôn ngữ C. . #include int chính () int i, n, tổng=0; printf("Nhập số bất kỳ. "); scanf("%d", &n); for(i=2; i<=n; i+=2) Làm thế nào để tìm tổng của n số chẵn đầu tiên trong C?Để tìm tổng của các số chẵn, chúng ta cần lặp qua các số chẵn từ 1 đến n. Khởi tạo một vòng lặp từ 2 đến N và tăng 2 trên mỗi lần lặp. Cấu trúc vòng lặp sẽ giống như for(i=2; i<=N; i+=2). Bên trong thân vòng lặp thêm giá trị trước đó của tổng với i i. e. tổng = tổng + tôi
Từ 1 đến n có bao nhiêu số chẵn?10 là số chẵn nhỏ nhất có hai chữ số. Có tổng 50 các số chẵn từ 1 đến 100. 2 là số nguyên tố chẵn duy nhất. Các số chia hết cho 2 là số chẵn.
Làm cách nào để in các số chẵn từ 1 đến 100 trong C?Viết chương trình C in tất cả các số chẵn từ 1 đến 100. . int main(){ int tôi; cho (i=1;i<=100;i++){ nếu(i%2==0){ printf("%d\n",i); |
