Finding n-th term of series 3, 13, 42, 108, 235…
Last Updated :
20 Feb, 2023
Given a number n, find the n-th term in the series 3, 13, 42, 108, 235…
Examples:
Input : 3
Output : 42
Input : 4
Output : 108
Constraints:
1 <= T <= 100
1 <= N <= 100
Naive Approach :
The series basically represents sums of natural numbers cube and number of terms multiplied by 2. The first term is the sum of the single number. The second term is the sum of two numbers, and so on.
Examples:
n = 2
2nd term equals to sum of 1st term and 8 i.e
A2 = A1 + 23 + n*2
= 1 + 8 + 4
= 13
Similarly,
A3 = A2 + 33 + n*2
= 9 + 27 + 6
= 42 and so on..
A simple solution is to add the first n natural numbers cube and number of terms multiplied by 2.
C++
#include <bits/stdc++.h>
using namespace std;
int magicOfSequence( int N)
{
int sum = 0;
for ( int i = 1; i <= N; i++)
sum += (i*i*i + i*2);
return sum;
}
int main()
{
int N = 4;
cout << magicOfSequence(N) << endl;
return 0;
}
|
Java
class GFG {
public static int magicOfSequence( int N)
{
int sum = 0 ;
for ( int i = 1 ; i <= N; i++)
sum += (i * i * i + i * 2 );
return sum;
}
public static void main(String args[])
{
int N = 4 ;
System.out.println(magicOfSequence(N));
}
}
|
Python3
def magicOfSequence(N) :
sum = 0
for i in range ( 1 , N + 1 ) :
sum + = (i * i * i + i * 2 )
return sum ;
N = 4
print (magicOfSequence(N))
|
C#
using System;
class GFG
{
public static int magicOfSequence( int N)
{
int sum = 0;
for ( int i = 1; i <= N; i++)
sum += (i * i * i + i * 2);
return sum;
}
static public void Main ()
{
int N = 4;
Console.WriteLine(magicOfSequence(N));
}
}
|
PHP
<?php
function magicOfSequence( $N )
{
$sum = 0;
for ( $i = 1; $i <= $N ; $i ++)
$sum += ( $i * $i * $i + $i *2);
return $sum ;
}
$N = 4;
echo magicOfSequence( $N );
?>
|
Javascript
<script>
function magicOfSequence( N)
{
let sum = 0;
for (let i = 1; i <= N; i++)
sum += (i*i*i + i*2);
return sum;
}
let N = 4;
document.write(magicOfSequence(N));
</script>
|
Time Complexity: O(n).
Auxiliary Space: O(1)
Efficient approach :
We know sum of cubes of first n natural numbers is (n*(n+1)/2)2. We also know that if we multiply i-th term by 2 and add all, we get sum of n terms as 2*n.
So our result is (n*(n+1)/2)2 + 2*n.
Example :
For n = 4 sum by the formula is
(4 * (4 + 1 ) / 2)) ^ 2 + 2*4
= (4 * 5 / 2) ^ 2 + 8
= (10) ^ 2 + 8
= 100 + 8
= 108
For n = 6, sum by the formula is
(6 * (6 + 1 ) / 2)) ^ 2 + 2*6
= (6 * 7 / 2) ^ 2 + 12
= (21) ^ 2 + 12
= 441 + 12
= 453
C++
#include <iostream>
using namespace std;
int magicOfSequence( int N)
{
return (N * (N + 1) / 2) + 2 * N;
}
int main()
{
int N = 6;
cout << magicOfSequence(N);
return 0;
}
|
Java
class GFG {
static int magicOfSequence( int N)
{
return (N * (N + 1 ) / 2 ) + 2 * N;
}
public static void main(String[] args)
{
int N = 6 ;
System.out.println(magicOfSequence(N));
}
}
|
Python 3
def magicOfSequence(N):
return (N * (N + 1 ) / 2 ) + 2 * N
N = 6
print ( int (magicOfSequence(N)))
|
C#
using System;
class GFG {
static int magicOfSequence( int N)
{
return (N * (N + 1) / 2) + 2 * N;
}
public static void Main()
{
int N = 6;
Console.Write(magicOfSequence(N));
}
}
|
PHP
<?php
function magicOfSequence( $N )
{
return ( $N * ( $N + 1) / 2) + 2 * $N ;
}
$N = 6;
echo magicOfSequence( $N ) . "\n" ;
?>
|
Javascript
<script>
function magicOfSequence( N)
{
return (N * (N + 1) / 2) + 2 * N;
}
let N = 6;
document.write(magicOfSequence(N));
</script>
|
Time Complexity: O(1)
Space Complexity: O(1) since constant space is used for variables
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