Given a series of natural numbers divided into groups as: (1, 2), (3, 4, 5, 6), (7, 8, 9, 10, 11, 12), (13, 14, 15, 16, 17, 18, 19, 20)….. and so on. Given a number N, the task is to find the sum of the numbers in the Nth group.
Examples:
Input : N = 3 Output : 57 Numbers in 3rd group are: 7, 8, 9, 10, 11, 12 Input : N = 10 Output : 2010
The first group has 2 terms,
the second group has 4 terms,
.
.
.
the nth group has 2n terms.
Now,
The last term of the first group is 2 = 1 × (1 + 1)
The last term of the second group is 6 = 2 × (2 + 1)
The last term of the third group is 12 = 3 × (3 + 1)
The last term of the fourth group is 20 = 4 × (4 + 1)
.
.
.
The last term of the nth group = n(n + 1).
Therefore, the sum of the numbers in the nth group is:
= sum of all the numbers upto nth group – sum of all the numbers upto (n – 1)th group
= [1 + 2 +……..+ n(n + 1)] – [1 + 2 +……..+ (n – 1 )((n – 1) + 1)]
=
=
=
=
![\frac{n(n+1)[n(n+1)+1]}{2} - \frac {n(n-1)[n(n-1)+1]}{2}](https://neveropen.co.uk/wp-content/uploads/2023/11/rubhabha_quicklatex.com-89692010b74eeedf1040c922a26cf372_l3-1.png "Rendered by QuickLaTeX.com")
![\frac{n[(n+1)(n(n+1)+1)-(n-1)(n(n-1)+1)]}{2}](https://neveropen.co.uk/wp-content/uploads/2023/11/neveropen_quicklatex.com-9d4d3f5f2cdf907704ac1373ef39d29d_l3.png "Rendered by QuickLaTeX.com")
![\frac{n[n[(n+1)^{2} - (n-1)^{2}]+2]}{2}](https://neveropen.co.uk/wp-content/uploads/2023/11/rubhabha_quicklatex.com-190f8b4ed3c1c0b2ebf2df8d67936cfa_l3-1.png "Rendered by QuickLaTeX.com")
Below is the implementation of the above approach:
C++
// C++ program to find sum in Nth group#include<bits/stdc++.h>using namespace std; //calculate sum of Nth groupint nth_group(int n){ return n * (2 * pow(n, 2) + 1);}//Driver codeint main(){ int N = 5; cout<<nth_group(N); return 0;} |
Java
// Java program to find sum// in Nth groupimport java.util.*;class GFG{// calculate sum of Nth groupstatic int nth_group(int n){ return n * (2 * (int)Math.pow(n, 2) + 1);}// Driver codepublic static void main(String arr[]){ int N = 5; System.out.println(nth_group(N));}}// This code is contributed by Surendra |
Python3
# Python program to find sum in Nth group# calculate sum of Nth groupdef nth_group(n): return n * (2 * pow(n, 2) + 1)# Driver codeN = 5print(nth_group(N)) |
C#
// C# program to find sum in Nth groupusing System; class gfg{ //calculate sum of Nth group public static double nth_group(int n) { return n * (2 * Math.Pow(n, 2) + 1); } //Driver code public static int Main() { int N = 5; Console.WriteLine(nth_group(N)); return 0; }}// This code is contributed by Soumik |
PHP
<?php// PHP program to find sum// in Nth group// calculate sum of Nth groupfunction nth_group($n){ return $n * (2 * pow($n, 2) + 1);}// Driver code$N = 5;echo nth_group($N);// This code is contributed // by jit_t?> |
Javascript
<script> // Javascript program to find sum in Nth group //calculate sum of Nth group function nth_group(n) { return n * (2 * Math.pow(n, 2) + 1); } let N = 5; document.write(nth_group(N)); </script> |
255
Time Complexity: O(1)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
