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++ 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 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 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# 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 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
?>

<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>

Sum = 77

Time Complexity: O(logn) 

Auxiliary Space: O(1)