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 

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

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

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

Time Complexity: O(logn) 

Auxiliary Space: O(1)