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+ // Project Euler Problem 6
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+ // Find the difference between the sum of the squares of the first n natural
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+ // numbers and the square of the sum of the first n natural numbers.
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+ //
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+ // Formula used:
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+ // Sum of squares = n(n+1)(2n+1)/6
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+ // Square of sum = (n(n+1)/2)^2
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+ //
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+ // Example:
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+ // Input : 100
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+ // Output: 25164150
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+
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+ #include <stdio.h>
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+
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+ // Function to compute sum of squares using formula
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+ long long sum_of_square (int n )
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+ {
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+ return (long long )n * (n + 1 ) * (2 * n + 1 ) / 6 ;
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+ }
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+
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+ // Function to compute square of sum using formula
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+ long long square_of_sum (int n )
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+ {
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+ long long sum = (long long )n * (n + 1 ) / 2 ;
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+ return sum * sum ;
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+ }
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+
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+ int main ()
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+ {
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+ int n ;
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+ scanf ("%d" , & n );
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+
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+ long long answer = square_of_sum (n ) - sum_of_square (n );
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+ printf ("%lld\n" , answer );
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+
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+ return 0 ;
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+ }
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+ //this calculates in o(1) ,enter the n till which u want to find. for our case (in question) its 100
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