Skip to content
Related Articles

Related Articles

Improve Article

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :