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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 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 using namespace std; // Function to calculate the sumint 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 codeint 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 sumstatic 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 termsdef 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 sumn = 5; # Find the Snprint ("Sum =", int(calculateSum(n))) # This code is contributed# by Surendra_Gangwar

## C#

 // C# program to find sum// of first n termsusing System; class GFG{ // Function to calculate the sumstatic 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 codepublic 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

 

## Javascript

 
Output:
Sum = 77

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