Given a series and a number n, the task is to find the sum of its first n terms. Below is the series:
2, 5, 13, 35, 97, …
Examples:
Input: N = 2 Output: 7 The sum of first 2 terms of Series is 2 + 5 = 7 Input: N = 4 Output: 55 The sum of first 4 terms of Series is 2 + 5 + 13 + 35 = 55
Approach: From this given series we find it is the sum of the Two GP series with common ratios 2, 3.
Sn = 2 + 5 + 13 + 35 + 97 … + upto nth term
Sn = (2^0 + 3^ 0) + (2^1 + 3^1) + (2^2 + 3^2) + (2^3 + 3^3)+ (2^4 + 3^4) …… + upto nth term
Sn = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 … + upto nth term) + ( 3^0 + 3^1 + 3^2 + 3^3 …… + unto nth term )
Since We know that the sum of n terms of the GP is given by the following formula:
Below is the implementation of the above approach:
C++
// C++ program for finding the sum// of first N terms of the series.#include <bits/stdc++.h>using namespace std;// CalculateSum function returns the final sumint calculateSum(int n){ // r1 and r2 are common ratios // of 1st and 2nd series int r1 = 2, r2 = 3; // a1 and a2 are common first terms // of 1st and 2nd series int a1 = 1, a2 = 1; return a1 * (pow(r1, n) - 1) / (r1 - 1) + a2 * (pow(r2, n) - 1) / (r2 - 1);}// Driver codeint main(){ int n = 4; // function calling for 4 terms cout << "Sum = " << calculateSum(n) << endl; return 0;} |
Java
//Java program for finding the sum//of first N terms of the series.public class GFG { //CalculateSum function returns the final sum static int calculateSum(int n) { // r1 and r2 are common ratios // of 1st and 2nd series int r1 = 2, r2 = 3; // a1 and a2 are common first terms // of 1st and 2nd series int a1 = 1, a2 = 1; return (int)(a1 * (Math.pow(r1, n) - 1) / (r1 - 1) + a2 * (Math.pow(r2, n) - 1) / (r2 - 1)); } //Driver code public static void main(String[] args) { int n = 4; // function calling for 4 terms System.out.println("Sum = " +calculateSum(n)); }} |
Python 3
# Python 3 program for finding the sum # of first N terms of the series. # from math import everythingfrom math import *# CalculateSum function returns the final sum def calculateSum(n) : # r1 and r2 are common ratios # of 1st and 2nd series r1, r2 = 2, 3 # a1 and a2 are common first terms # of 1st and 2nd series a1, a2 = 1, 1 return (a1 * (pow(r1, n) - 1) // (r1 - 1) + a2 * (pow(r2, n) - 1) // (r2 - 1))# Driver Codeif __name__ == "__main__" : n = 4 # function calling for 4 terms print("SUM = ",calculateSum(n))# This code is contributed by ANKITRAI1 |
C#
// C# program for finding the sum// of first N terms of the series.using System;class GFG {// CalculateSum function // returns the final sumstatic int calculateSum(int n){// r1 and r2 are common ratios// of 1st and 2nd seriesint r1 = 2, r2 = 3;// a1 and a2 are common first // terms of 1st and 2nd seriesint a1 = 1, a2 = 1;return (int)(a1 * (Math.Pow(r1, n) - 1) / (r1 - 1) + a2 * (Math.Pow(r2, n) - 1) / (r2 - 1));}// Driver codestatic public void Main (){ int n = 4; // function calling for 4 terms Console.Write("Sum = " + calculateSum(n));}}// This code is contributed by Raj |
PHP
<?php// PHP program for finding the sum // of first N terms of the series. // CalculateSum function returns // the final sum function calculateSum($n) { // r1 and r2 are common ratios // of 1st and 2nd series $r1 = 2; $r2 = 3; // a1 and a2 are common first // terms of 1st and 2nd series $a1 = 1; $a2 = 1; return $a1 * (pow($r1, $n) - 1) / ($r1 - 1) + $a2 * (pow($r2, $n) - 1) / ($r2 - 1); } // Driver code $n = 4; // function calling for 4 terms echo "Sum = ", calculateSum($n); // This code is contributed by ash264?> |
Javascript
<script>//javascript program for finding the sum//of first N terms of the series.//CalculateSum function returns the final sumfunction calculateSum(n){ // r1 and r2 are common ratios // of 1st and 2nd series var r1 = 2, r2 = 3; // a1 and a2 are common first terms // of 1st and 2nd series var a1 = 1, a2 = 1; return parseInt((a1 * (Math.pow(r1, n) - 1) / (r1 - 1) + a2 * (Math.pow(r2, n) - 1) / (r2 - 1)));} //Driver code var n = 4; // function calling for 4 terms document.write("Sum = " +calculateSum(n));// This code contributed by shikhasingrajput </script> |
Output
Sum = 55
Time Complexity: O(log n)
Auxiliary Space: O(1), As constant extra space is used.
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