程序打印给定第n个项的总和

给定n的值。您必须找到序列的总和，其中序列的第n^个项由下式给出：
T _n = n ² –(n – 1) ²
例子 ：

Input : 3
Output : 9

Explanation: So here the tern of the sequence upto n = 3 are: 
             1, 3, 5 And hence the required sum is = 1 + 3 + 5 = 9

Input : 6
Output : 36

简单方法
只需使用一个循环并计算每个项的总和并打印总和即可。

C++

// CPP program to find summation of series
#include 
using namespace std;
 
int summingSeries(long n)
{
    // use of loop to calculate
    // sum of each term
    int S = 0;
    for (int i = 1; i <= n; i++)
        S += i * i - (i - 1) * (i - 1);
     
    return S;
}
 
// Driver Code
int main()
{
    int n = 100;
    cout << "The sum of n term is: "
        << summingSeries(n) << endl;
    return 0;
}

Java

// JAVA program to find summation of series
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
 
class GFG
{
 
    // function to calulate sum of series
    static int summingSeries(long n)
    {
        // use of loop to calculate
        // sum of each term
        int S = 0;
        for (i = 1; i <= n; i++)
            S += i * i - (i - 1) * (i - 1);    
         
        return S;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.println("The sum of n term is: " +
                            summingSeries(n));
    }
}

Python3

# Python3 program to find summation
# of series
 
def summingSeries(n):
 
    # use of loop to calculate
    # sum of each term
    S = 0
    for i in range(1, n+1):
        S += i * i - (i - 1) * (i - 1)
     
    return S
 
# Driver Code
n = 100
print("The sum of n term is: ",
           summingSeries(n), sep = "")
# This code is contributed by Smitha.

C#

// C# program to illustrate...
// Summation of series
using System;
 
class GFG
{
 
    // function to calculate sum of series
    static int summingSeries(long n)
    {
 
        // Using the pow function calculate
        // the sum of the series
        return (int)Math.Pow(n, 2);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 100;
        Console.Write("The sum of n term is: " +
                        summingSeries(n));
    }
}
 
// This code contribute by Parashar...

Javascript

// Javascript program to find summation of series
 
function summingSeries(n)
{
    // use of loop to calculate
    // sum of each term
    let S = 0;
    for (let i = 1; i <= n; i++)
        S += i * i - (i - 1) * (i - 1);
     
    return S;
}
 
// Driver Code
let n = 100;
document.write("The sum of n term is: " + summingSeries(n));
 
// This code is contributed by rishavmahato348.

输出：

The sum of n term is: 10000

时间复杂度– O(N)
空间复杂度– O(1)
高效方法
使用数学方法可以更有效地解决此问题。

T_n = n² – (n-1)²
Sum of the series is given by (S) = SUM( T_n )
LET US TAKE A EXAMPLE IF 
N = 4 
It means there should be 4 terms in the series so
1^st term = 1² – ( 1 – 1 )² 
2^nd term = 2² – ( 2 – 1 )² 
3^th term = 3² – ( 3 – 1 )² 
4^th term = 4² – ( 3 – 1 )²
SO SUM IS GIVEN BY = (1 – 0) + (4 – 1) + (9 – 4) + (16 – 9) 
= 16
FROM THIS WE HAVE NOTICE THAT 1, 4, 9 GET CANCELLED FROM THE SERIES 
ONLY 16 IS LEFT WHICH IS EQUAL TO THE SQUARE OF N
So from the above series we notice that each term gets canceled from the next term, only the last term is left which is equal to N².
 

C++

// CPP program to illustrate...
// Summation of series
 
#include 
using namespace std;
 
int summingSeries(long n)
{
    // Sum of n terms is n^2
    return pow(n, 2);
}
 
// Driver Code
int main()
{
    int n = 100;
    cout << "The sum of n term is: "
         << summingSeries(n) << endl;
    return 0;
}

Java

// JAVA program to illustrate...
// Summation of series
 
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
 
class GFG
{
 
    // function to calculate sum of series
    static int summingSeries(long n)
    {
 
        // Using the pow function calculate
        // the sum of the series
        return (int)Math.pow(n, 2);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.println("The sum of n term is: " +
                            summingSeries(n));
    }
}

Python3

# Python3 program to illustrate...
# Summation of series
import math
 
def summingSeries(n):
 
    # Sum of n terms is  n^2
    return math.pow(n, 2)
 
# Driver Code
n = 100
print ("The sum of n term is: ",
        summingSeries(n))
# This code is contributed by mits.

C＃

// C# program to illustrate...
// Summation of series
using System;
 
class GFG
{
    // function to calculate sum of series
    static int summingSeries(long n)
    {
        // Using the pow function calculate
        // the sum of the series
        return (int)Math.Pow(n, 2);
    }
 
    // Driver code
    public static void Main()
    {
        int n = 100;
           Console.Write("The sum of n term is: " +
                              summingSeries(n));
    }
}
 
// This code is contributed by nitin mittal.

Java脚本

// Javascript program to illustrate...
// Summation of series
 
function summingSeries(n)
{
    // Sum of n terms is n^2
    return Math.pow(n, 2);
}
 
// Driver Code
let n = 100;
document.write("The sum of n term is: "
    + summingSeries(n) + "<br>");
     
    // This code is contributed by subham348.

输出：

The sum of n term is: 10000

时间复杂度– O(1)
空间复杂度– O(1)