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Yash Agarwal

For proving that n(n+1)/2 is an integer, you take if cases. If n is o mod 2, then it would obviously be an integer since anything multiplied by 0 is 0. But, if n is 1 mod 2, then (n+1) is equal to 2 mod 2 which is equal to 0 mod 2, making the division by 2 an integer.

For proving that n(n+1)(2n+1)/6 is an integer, we first have to take cases and find a pattern. Let’s suppose n mod 6 is equal to 0. Then the rest would obviously be divisible by 6, making it an integer. But, if n = 1 mod 6, then (n+1) is equal to 2 mod 6 and 2n+1 is equal to 3 mod 3. 1*2*3 mod 6 = 6 mod 6 = 0 mod 6, which means it would be divisible by 6, making it an integer. Similarly, if n equals 2 mod 6, then n+1 equal to 3 mod 6 and 2n+1 is equal to 5 mod 6. 2*3*5 mod 6 = 30 mod 6 = 0 mod 6, which means that it would be an integer. It is the same for all values of n up to 5. They all make 0 mod 6, which means that it will always be an integer!

** This response is written by Yash Agarwal.**