Sum of the series 3, 20, 63, 144, ……
Last Updated :
31 Aug, 2022
Find the sum of first n terms of the given series:
3, 20, 63, 144, .....
Examples:
Input : n = 2
Output : 23
Input : n =4
Output : 230
Approach:
First, we have to find the general term (Tn) of the given series.
series can we written in the following way also:
(3 * 1^2), (5 * 2^2), (7 * 3^2), (9 * 4^2), .......up t n terms
Tn = (General term of series 3, 5, 7, 9 ....) X (General term of series 1^2, 2^2, 3^2, 4^2 ....)
Tn = (3 + (n-1) * 2) X ( n^2 )
Tn = 2*n^3 + n^2
We can write the sum of the series in the following ways:
Sn = 3 + 20 + 63 + 144 + ........up to the n terms
[Tex]$$
Sn = 2 \times \sum_{n=1}^{n} n^{3} + \sum_{n=1}^{n} n^{2}
$$[/Tex]Sn = 2 * (sum of n terms of n^3 ) + (sum of n terms of n^2)
Following are the formulas of sum of n terms of n^3 and n^2 :
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int calculateSum( int n)
{
return (2 * pow ((n * (n + 1) / 2), 2)) +
((n * (n + 1) * (2 * n + 1)) / 6);
}
int main()
{
int n = 4;
cout << "Sum = " << calculateSum(n) << endl;
return 0;
}
|
Java
import java.io.*;
public class GFG
{
static int calculateSum( int n)
{
return ( int )(( 2 * Math.pow((n * (n + 1 ) / 2 ), 2 ))) +
((n * (n + 1 ) * ( 2 * n + 1 )) / 6 );
}
public static void main (String[] args) {
int n = 4 ;
System.out.println( "Sum = " + calculateSum(n));
}
}
|
Python3
def calculateSum(n):
return (( 2 * (n * (n + 1 ) / 2 ) * * 2 ) +
((n * (n + 1 ) * ( 2 * n + 1 )) / 6 ))
n = 4
print ( "Sum =" ,calculateSum(n))
|
C#
using System;
class GFG
{
static int calculateSum( int n)
{
return ( int )((2 * Math.Pow((n * (n + 1) / 2), 2))) +
((n * (n + 1) * (2 * n + 1)) / 6);
}
public static void Main ()
{
int n = 4;
Console.WriteLine( "Sum = " + calculateSum(n));
}
}
|
PHP
<?php
function calculateSum( $n )
{
return (2 * pow(( $n * ( $n + 1) / 2), 2)) +
(( $n * ( $n + 1) * (2 * $n + 1)) / 6);
}
$n = 4;
echo "Sum = " , calculateSum( $n );
?>
|
Javascript
<script>
function calculateSum(n)
{
return parseInt(((2 * Math.pow((n * (n + 1) / 2), 2))) +
((n * (n + 1) * (2 * n + 1)) / 6));
}
var n = 4;
document.write( "Sum = " + calculateSum(n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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