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/*
The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum
(n(n+1)/2)^2 - n(n+1)(2n+1)/6 = n(n+1)/12 ( 3n*n + 3n - 4n -2)
= n(n+1)/12 ( 3n*n -n -2)
*/
#include<stdio.h>
typedef unsigned long long int uint64;
int main()
{
int n;
printf("Enter number : ");
scanf("%d",&n);
printf("%llu\n",(uint64) (n*(n+1)*(3*n*n -n -2))/12 );
return 0;
}