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 :
![$$ \sum_{n=1}^{n} n^{3} = \left[\frac{n \times \big(n + 1 \big) }{2} \right]^{2} $$ $$ \sum_{n=1}^{n} n^{2} = \frac{n \times \big(n + 1 \big) \times \big(2*n + 1 \big) }{6} $$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-799eea18da3634818b29530539e24b1f_l3.png "Rendered by QuickLaTeX.com")
![Total = 2 \times \left[\frac{n \times \big(n + 1 \big) }{2} \right]^{2} + \frac{n \times \big(n + 1 \big) \times \big(2*n + 1 \big) }{6}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-91e6be86531e8a55afa06e36a30aaa45_l3.png "Rendered by QuickLaTeX.com")
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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