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Program to print the sum of the given nth term
  • Last Updated : 08 Jul, 2020

Given the value of the n. You have to find the sum of the series where the nth term of the sequence is given by:

Tn = n2 – ( n – 1 )2

Examples :

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

Simple Approach
Just use a loop and calculate the sum of each term and print the sum.

C++

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// CPP program to find summation of series
#include <bits/stdc++.h>
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;
}

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Java

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// 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));
    }
}

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Python3

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# 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.

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C#

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// 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...

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PHP

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<?php
// PHP program to find 
// summation of series
  
function summingSeries( $n)
{
      
    // use of loop to calculate
    // sum of each term
    $S = 0; 
    for ($i = 1; $i <= $n; $i++) 
         $S += $i * $i - ($i - 1) * 
                       ($i - 1);
      
    return $S;
}
  
// Driver Code
$n = 100;
echo "The sum of n term is: ",
summingSeries($n) ;
  
// This code contribute by vt_m.
?>

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Output:



The sum of n term is: 10000

Time complexity – O(N)
Space complexity – O(1)

Efficient Approach
Use of mathematical approach can solve this problem in more efficient way.

Tn = n2 – (n-1)2

Sum of the series is given by (S) = SUM( Tn )

LET US TAKE A EXAMPLE IF
N = 4
It means there should be 4 terms in the series so

1st term = 12 – ( 1 – 1 )2
2nd term = 22 – ( 2 – 1 )2
3th term = 32 – ( 3 – 1 )2
4th 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++

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// CPP program to illustrate...
// Summation of series
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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));
    }
}

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?php
// PHP program to illustrate...
// Summation of series
  
function summingSeries($n)
{
    // Sum of n terms is  n^2
    return pow($n, 2);
}
  
// Driver Code
$n = 100;
echo "The sum of n term is: ",
summingSeries($n);
  
// This code contribute by vt_m.
?>

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Output:

The sum of n term is: 10000


Time complexity – O(1)
Space complexity – O(1)

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