Write a program in java to compute the following expressions.
S = 1 * ( 1+2) *(1+2+3) * (1+2+3+4)+ ......n
terms
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Answers
Answer:
Input : n = 5
Output : 35
Explanation :
(1) + (1+2) + (1+2+3) + (1+2+3+4) + (1+2+3+4+5) = 35
Input : n = 10
Output : 220
Explanation :
(1) + (1+2) + (1+2+3) + .... +(1+2+3+4+.....+10) = 220
Recommended: Please solve it on PRACTICE first, before moving on to the solution.
Naive Approach :
Below is implementation of above series :
C++
Java
// JAVA Code For Sum of the series
import java.util.*;
class GFG {
// Function to find sum of given series
static int sumOfSeries(int n)
{
int sum = 0;
for (int i = 1 ; i <= n ; i++)
for (int j = 1 ; j <= i ; j++)
sum += j;
return sum;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 10;
System.out.println(sumOfSeries(n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python
C#
PHP
Output :
220
Efficient Approach :
Let n^{th} term of the series 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)…(1 + 2 + 3 +..n) be denoted as an
an = Σn1 i = \frac{n (n + 1)}{2} = \frac{(n^2 + n)}{2}
Sum of n-terms of series
Σn1 an = Σn1 \frac{(n^2 + n)}{2}
= \frac{1}{2} Σ [ n^2 ] + Σ [ n ]
= \frac{1}{2} * \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{2} * \frac{n(n+1)}{2}
= \frac{n(n+1)(2n+4)}{12}
Below is implementation of above approach :
C++
// CPP program to find sum of given series
#include <bits/stdc++.h>
using namespace std;
// Function to find sum of given series
int sumOfSeries(int n)
{
return (n * (n + 1) * (2 * n + 4)) / 12;
}
// Driver Function
int main()
{
int n = 10;
cout << sumOfSeries(n);
}
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