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Sum of first n terms of a given series 3, 6, 11, …..
  • Difficulty Level : Easy
  • Last Updated : 25 Feb, 2021

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++




// 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;
}


Java




// 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.


Python3




# 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


C#




// 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


PHP




<?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
?>


Javascript




<script>
 
// Javascript program to find sum of first n terms
// Function to calculate the sum
 
function calculateSum(n)
{
    // starting number
    let a1 = 1, a2 = 2;
 
    // Common Ratio
    let r = 2;
 
    // Common difference
    let d = 1;
 
    return (n) * (2 * a1 + (n - 1) * d) / 2 + a2
            * (Math.pow(r, n) - 1) / (r - 1);
}
 
// Driver code
 
    // N th term to be find
    let n = 5;
 
    // find the Sn
    document.write("Sum = " + calculateSum(n));
 
 
 
// This code is contributed by Mayank Tyagi
 
</script>


Output: 

Sum = 77

 

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