- Related Questions & Answers
- Sum of sum of first n natural numbers in C++
- C Program for cube sum of first n natural numbers?
- C++ Program for cube sum of first n natural numbers?
- Sum of squares of first n natural numbers in C Program?
- C++ Program for Sum of squares of first n natural numbers?
- Program to find sum of first n natural numbers in C++
- Program for cube sum of first n natural numbers in C++
- C Program for the cube sum of first n natural numbers?
- Python Program for cube sum of first n natural numbers
- Java Program to cube sum of first n natural numbers
- Python Program for Sum of squares of first n natural numbers
- Sum of square-sums of first n natural numbers
- 8085 program to find the sum of first n natural numbers
- Java Program to calculate Sum of squares of first n natural numbers
- PHP program to calculate the sum of square of first n natural numbers

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

The concept of finding the sum of sum of integers is found such that first, we will find the sum of numbers up to n and then add all the sums to get a value which will be the sum of sum which is our desired sum.

For this problem, we are given a number n up to which we have to find the sum of the sum. Let's take an example to find this sum.

n = 4

Now we will find the sum of numbers for every number from 1 to 4 :

Sum of numbers till 1 = 1 Sum of numbers till 2 = 1 + 2 = 3 Sum of numbers till 3 = 1 + 2 + 3 = 6 Sum of numbers till 4 = 1 + 2 + 3 + 4 = 10 Now we will find the sum of sum of numbers til n : Sum = 1+3+6+10 = 20

For finding the sum of sum of n natural number, we have two methods :

**Method 1** − Using the for loops (inefficient)

**Method 2** − Using the mathematical formula (efficient)

In this method, we will use two for loops to find the sum of sum. The inner loop finds the sum of natural number and outer loop adds this sum to sum2 and increases the number by one.

#include <stdio.h> int main() { int n = 4; int sum=0, s=0; for(int i = 1; i< n; i++){ for(int j= 1; j<i;j++ ){ s+= j; } sum += s; } printf("the sum of sum of natural number till %d is %d", n,sum); return 0; }

The sum of sum of natural number till 4 is 5

We have a mathematical formula for finding the sum of sum of n natural numbers. The mathematical formula method is an efficient method.

The mathematical formula to find the sum of sum of n natural number :

sum = n*(n+1)*(n+2)/2

#include <stdio.h> int main() { int n = 4; int sum = (n*(n+1)*(n+2))/2; printf("the sum of sum of natural number till %d is %d", n,sum); return 0; }

the sum of sum of natural number till 4 is 60

Advertisements