Sum of first n terms of a given series 3, 6, 11, …..

Given a series and a number n, the task is to find the sum of its first n terms. Below is the series:

3, 6, 11, 20, ….

Examples:

Input: N = 2
Output: 9
The sum of first 2 terms of Series is
3 + 6 = 9

Input: N = 3
Output: 20
The sum of first 3 terms of Series is
3 + 6 + 11 = 20


Approach: This problem can easily solved by observing that the nth term of the series :

Sn = 3 + 6 + 11 + 20 … + upto nth term
Sn = (1 + 2^1) + (2 + 2^2) + (3 + 2^3)+ (4 + 2^4) …… + upto nth term
Sn = (1 + 2 + 3 + 4 …. + upto nth term) + ( 2^1 + 2^2 + 2^3 …… + unto nth term )

We observe that Sn is a summation of two series: AP and GP
As we know the sum of first n terms of AP is given by

    $$S_n=\frac{n}{2} \left(2 \times a1+(n-1) \times d\right)$$

And also the Sum of first n terms of G.P is given by

    $$Sn=a2 \times \left(\frac{r^n-1}{r-1}\right)$$

Hence the total sum is given by sum of both AP and GP.

     $$Total=\frac{n}{2} \left(2 \times a1+(n-1) \times d\right) +a2 \times \left(\frac{r^n-1}{r-1}\right)$$

Below is the implementation of above approach.

C++

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// C++ program to find sum of first n terms
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate the sum
int calculateSum(int n)
{
    // starting number
    int a1 = 1, a2 = 2;
  
    // Common Ratio
    int r = 2;
  
    // Common difference
    int d = 1;
  
    return (n) * (2 * a1 + (n - 1) * d) / 2 + a2 
               * (pow(r, n) - 1) / (r - 1);
}
  
// Driver code
int main()
{
  
    // N th term to be find
    int n = 5;
  
    // find the Sn
    cout << "Sum = " << calculateSum(n);
  
    return 0;
}

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Java

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// Java program to find sum of first n terms
  
import java.io.*;
  
class GFG {
  
// Function to calculate the sum
static int calculateSum(int n)
{
    // starting number
    int a1 = 1, a2 = 2;
  
    // Common Ratio
    int r = 2;
  
    // Common difference
    int d = 1;
  
    return (n) * (2 * a1 + (n - 1) * d) / 2 + a2 
            * (int)(Math.pow(r, n) - 1) / (r - 1);
}
  
// Driver code
    public static void main (String[] args) {
        // N th term to be find
    int n = 5;
  
    // find the Sn
    System.out.print( "Sum = " + calculateSum(n));
    }
}
// This code is contributed by inder_verma.

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Python3

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# Python3 program to find 
# sum of first n terms
def calculateSum(n):
    # First term of AP
    a1 = 1;
      
    # First term of GP
    a2 = 2;
      
    # common ratio of GP
    r = 2;
      
    # common difference Of AP
    d = 1;
    return ((n) * (2 * a1 + (n - 1) * d) / 
                   2 + a2 * (pow(r, n) - 1) / 
                  (r - 1));
  
# Driver Code
  
# no. of the terms 
# for the sum
n = 5;
  
# Find the Sn
print ("Sum =", int(calculateSum(n)))
  
# This code is contributed 
# by Surendra_Gangwar

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C#

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// C# program to find sum
// of first n terms
using System;
  
class GFG 
{
  
// Function to calculate the sum
static int calculateSum(int n)
{
    // starting number
    int a1 = 1, a2 = 2;
  
    // Common Ratio
    int r = 2;
  
    // Common difference
    int d = 1;
  
    return (n) * (2 * a1 + (n - 1) * d) / 2 + a2 * 
             (int)(Math.Pow(r, n) - 1) / (r - 1);
}
  
// Driver code
public static void Main () 
{
    // N th term to be find
    int n = 5;
      
    // find the Sn
    Console.WriteLine("Sum = " + calculateSum(n));
}
}
  
// This code is contributed 
// by inder_verma

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PHP

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<?php
// PHP program to find sum of first n terms
  
// Function to calculate the sum
function calculateSum($n)
{
    // starting number
    $a1 = 1;
    $a2 = 2;
      
    // Common Ratio
    $r = 2;
      
    // Common difference
    $d = 1;
      
    return ($n) * (2 * $a1 + ($n - 1) * $d) / 2 + 
            $a2 * (pow($r, $n) - 1) / ($r - 1);
}
  
// Driver code
  
// Nth term to be find
$n = 5;
  
// find the Sn
echo "Sum = ", calculateSum($n);
  
// This code is contributed 
// by Shashank_Sharma
?>

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Output:

Sum = 77


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