Given an integer N. The task is to find the sum of the first N terms of the series 5, 11, 19, 29, 41, . . . till Nth term.
Examples:
Input: N = 5
Output: 105
Explanation: 5 + 11 + 19 + 29 + 41 = 105.Input: N = 2
Output: 16
Explanation: The terms are 5 and 11
Approach: From the given series first determine the Nth term:
1st term = 5 = 1 + 4 = 1 + 22
2nd term = 11 = 2 + 9 = 2 + 32
3rd term = 19 = 3 + 16 = 3 + 42
4th term = 29 = 4 + 25 = 4 + 52
.
.
Nth term = N + (N+1)2
So the Nth term can be written as: TN = N + (N+1)2
Therefore the sum up to N terms becomes
1 + 22 + 2 + 32 + 3 + 42 + . . . + N + (N+1)2
= [1 + 2 + 3 + . . . + N] + [22 + 32 + 42 + . . . + (N+1)2]
= (N*(N+1))/2 + [(N+1)*(N+2)*(2*N + 3)]/6 – 1
= [N*(N+2)*(N+4)]/3
Therefore the sum of the first N terms can be given as: SN = [N*(N+2)*(N+4)]/3
Illustration:
For example, take N = 5
The output will be 105.
Use N = 5, then N*(N+2)*(N+4)/3
= 5 * 7 * 9/3 = 5 * 7 * 3 = 105.
This is same as 5 + 11 + 19 + 29 + 41
Below is the implementation of the above approach.
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate// the sum of first N termsint nthSum(int N){ // Formula for sum of N terms int ans = (N * (N + 2) * (N + 4)) / 3; return ans;}// Driver codeint main(){ int N = 5; cout << nthSum(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;public class GFG{ // Function to calculate // the sum of first N terms static int nthSum(int N) { // Formula for sum of N terms int ans = (N * (N + 2) * (N + 4)) / 3; return ans; } // Driver code public static void main(String args[]) { int N = 5; System.out.println(nthSum(N)); }}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python code to implement the above approach# Function to calculate# the sum of first N termsdef nthSum(N): # Formula for sum of N terms ans = (int)((N * (N + 2) * (N + 4)) / 3) return ans# Driver codeN = 5print(nthSum(N))# This code is contributed by Taranpreet |
C#
// C# program for the above approachusing System;class GFG{ // Function to calculate// the sum of first N termsstatic int nthSum(int N){ // Formula for sum of N terms int ans = (N * (N + 2) * (N + 4)) / 3; return ans;}// Driver codepublic static void Main(){ int N = 5; Console.Write(nthSum(N));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to calculate // the sum of first N terms function nthSum(N) { // Formula for sum of N terms let ans = (N * (N + 2) * (N + 4)) / 3; return ans; } // Driver code let N = 5; document.write(nthSum(N)); // This code is contributed by Potta Lokesh</script> |
105
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.
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