Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+n) in C++

In this problem, we are given a number n. Our task is to create a program to find the sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + … + (1+2+3+4+...+n).

Lets example to understand the problem,

Input

n = 4

Output

Explanation −(1) + (1+2) + (1+2+3) + (1+2+3+4) = 20

A simple solution to the problem will be creating the series by using two loops.

Algorithm

Initialize sum = 0
Step 1: Loop for i -> 1 to n i.e i = 1 to i <= n.
   Step 1.1: Loop for j -> 1 to i i.e. i = 1 to i <= i.
   Step 1.1.1: update sum i.e. sum += j.
Step 2: return sum.

Example

Program to illustrate the working of our solution,

Live Demo

#include <iostream>
using namespace std;
int calcSeriesSum(int n) {
   int sum = 0;
   for (int i = 1 ; i <= n ; i++)
      for (int j = 1 ; j <= i ; j++)
         sum += j;
   return sum;
}
int main() {
   int n = 7;
   cout<<"Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+"<<n<<") is "<<calcSeriesSum(n);
   return 0;
}

Output

Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+7) is 84

But this approach is not effective.

An effective solution could be deriving the general formula for finding the sum of the series.

sum = 1 + (1+2) + (1+2+3) + (1+2+3+4) …
sum = ∑ ( (1+2+3+4+5+...) )
sum = ∑ ( n(n+1)/2)
sum = ½ ∑ ( n^2 + n) = ½ (∑ (n2) + ∑ n)
sum = ½ [ (n(n+1)(2n+1))/6 ) + ½ ( n(n+1)/2 ]
sum = ½ [ (n(n+1))/2 ( (2n+1)/3 + 1) ]
sum = ½ [ ((n(n+1))/2) * (2n + 1 + 3)/3 ]
sum = ½ [ (n(n+1)(2n+4))/6]
sum = (n(n + 1)(2n + 4))/6

Example

Program to illustrate the working of our solution,

Live Demo

#include <iostream>
using namespace std;
int calcSeriesSum(int n) {
   return (n*(n + 1)*(2*n + 4))/12;
}
int main() {
int n = 7;
   cout<<"Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+"<<n<<") is "<<calcSeriesSum(n);
}

Output

Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ... + (1+2+3+4+...+7) is 84

Sudhir sharma

Published on 14-Aug-2020 13:35:25

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