给定正整数n 。问题是找到给定序列2 +(2 + 4)+(2 + 4 + 6)+(2 + 4 + 6 + 8)+……+(2 + 4 + 6 + 8 +…的总和。 。+ 2n) ,其中系列中的第i个项是第一个i个偶数之和。
例子:
Input : n = 2
Output : 8
(2) + (2+4) = 8
Input : n = 5
Output : 70
(2) + (2+4) + (2+4+6) + (2+4+6+8) + (2+4+6+8+10) = 70
天真的方法:使用两个循环获得每个第i个项的总和,然后将这些总和加到最终总和上。
C++
// C++ implementation to find the sum
// of the given series
#include
using namespace std;
// function to find the sum
// of the given series
int sumOfTheSeries(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++) {
// first term of each i-th term
int k = 2;
for (int j = 1; j <= i; j++) {
sum += k;
// next term
k += 2;
}
}
// required sum
return sum;
}
// Driver program to test above
int main()
{
int n = 5;
cout << "Sum = "
<< sumOfTheSeries(n);
return 0;
}
Java
// Java implementation to find the
// sum of the given series
class GFG{
// function to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++) {
// first term of each i-th term
int k = 2;
for (int j = 1; j <= i; j++) {
sum += k;
// next term
k += 2;
}
}
// required sum
return sum;
}
// Driver program to test above
public static void main(String[] args)
{
int n = 5;
System.out.printf("Sum = %d",
sumOfTheSeries(n));
}
}
// This code is contriubted by
// Smitha Dinesh Semwal
Python3
# Python3 implementation to find
# the sum of the given series
# function to find the sum
# of the given series
def sumOfTheSeries(n):
sum = 0
for i in range(0, n + 1):
# first term of each i-th
# term
k = 2
for j in range(1, i + 1):
sum = sum + k;
# next term
k = k + 2
# required sum
return sum;
# Driver program to test above
n = 5
ans = sumOfTheSeries(n);
print (ans)
# This code is contributed by saloni1297.
C#
// C# implementation to find the
// sum of the given series
using System;
class GFG{
// function to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++) {
// first term of each i-th term
int k = 2;
for (int j = 1; j <= i; j++) {
sum += k;
// next term
k += 2;
}
}
// required sum
return sum;
}
// Driver program to test above
public static void Main()
{
int n = 5;
Console.Write("Sum = "+
sumOfTheSeries(n));
}
}
// This code is contriubted by
// vt_m
PHP
Javascript
C++
// C++ implementation to find the sum
// of the given series
#include
using namespace std;
// functionn to find the sum
// of the given series
int sumOfTheSeries(int n)
{
// sum of 1st n natural numbers
int sum_n = (n * (n + 1) / 2);
// sum of squares of 1st n natural numbers
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
// required sum
return (sum_n + sum_sq_n);
}
// Driver program to test above
int main()
{
int n = 5;
cout << "Sum = "
<< sumOfTheSeries(n);
return 0;
}
Java
// Java implementation to find the
// sum of the given series
class GFG{
// functionn to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
// sum of 1st n natural numbers
int sum_n = (n * (n + 1) / 2);
// sum of squares of 1st n natural
// numbers
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
// required sum
return (sum_n + sum_sq_n);
}
// Driver program to test above
public static void main(String[] args)
{
int n = 5;
System.out.printf("Sum = %d",
sumOfTheSeries(n));
}
}
// This code is contriubted by
//Smitha Dinesh Semwal
Python3
# Python3 implementation to find
# the sum of the given series
# functionn to find the sum
# of the given series
def sumOfTheSeries(n):
# sum of 1st n natural numbers
sum_n = int((n * (n + 1) / 2));
# sum of squares of 1st n natural numbers
sum_sq_n = int ((n * (n + 1) / 2) * (2 * n + 1) / 3)
# required sum
return (sum_n + sum_sq_n);
# Driver program to test above
n = 5
ans = sumOfTheSeries(n)
print (ans)
# This code is contributed by saloni1297.
C#
// C# implementation to find the
// sum of the given series
using System;
class GFG{
// functionn to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
// sum of 1st n natural numbers
int sum_n = (n * (n + 1) / 2);
// sum of squares of 1st n
// natural numbers
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
// required sum
return (sum_n + sum_sq_n);
}
// Driver program to test above
public static void Main()
{
int n = 5;
Console.Write("Sum = "+
sumOfTheSeries(n));
}
}
// This code is contriubted by
// vt_m
PHP
Javascript
输出:
Sum = 70
高效方法:
令n为给定级数的第n个项。
an = (2 + 4 + 6 + 8 +....+ 2n).
= sum of first n even numbers.
= n * (n + 1).
= n2 + n.
有关以上公式的证明,请参阅此帖子。
现在,
有关以上公式的证明,请参考此和这篇文章。
C++
// C++ implementation to find the sum
// of the given series
#include
using namespace std;
// functionn to find the sum
// of the given series
int sumOfTheSeries(int n)
{
// sum of 1st n natural numbers
int sum_n = (n * (n + 1) / 2);
// sum of squares of 1st n natural numbers
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
// required sum
return (sum_n + sum_sq_n);
}
// Driver program to test above
int main()
{
int n = 5;
cout << "Sum = "
<< sumOfTheSeries(n);
return 0;
}
Java
// Java implementation to find the
// sum of the given series
class GFG{
// functionn to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
// sum of 1st n natural numbers
int sum_n = (n * (n + 1) / 2);
// sum of squares of 1st n natural
// numbers
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
// required sum
return (sum_n + sum_sq_n);
}
// Driver program to test above
public static void main(String[] args)
{
int n = 5;
System.out.printf("Sum = %d",
sumOfTheSeries(n));
}
}
// This code is contriubted by
//Smitha Dinesh Semwal
Python3
# Python3 implementation to find
# the sum of the given series
# functionn to find the sum
# of the given series
def sumOfTheSeries(n):
# sum of 1st n natural numbers
sum_n = int((n * (n + 1) / 2));
# sum of squares of 1st n natural numbers
sum_sq_n = int ((n * (n + 1) / 2) * (2 * n + 1) / 3)
# required sum
return (sum_n + sum_sq_n);
# Driver program to test above
n = 5
ans = sumOfTheSeries(n)
print (ans)
# This code is contributed by saloni1297.
C#
// C# implementation to find the
// sum of the given series
using System;
class GFG{
// functionn to find the sum
// of the given series
static int sumOfTheSeries(int n)
{
// sum of 1st n natural numbers
int sum_n = (n * (n + 1) / 2);
// sum of squares of 1st n
// natural numbers
int sum_sq_n = (n * (n + 1) / 2) *
(2 * n + 1) / 3;
// required sum
return (sum_n + sum_sq_n);
}
// Driver program to test above
public static void Main()
{
int n = 5;
Console.Write("Sum = "+
sumOfTheSeries(n));
}
}
// This code is contriubted by
// vt_m
的PHP
Java脚本
输出:
Sum = 70