"A Mathematician's Apology"

Relevant wiki: Sum of n, n², or n³

The general formula for the sum of the first cubes is $\left[ \frac { n(n+1) }{ 2 } \right] ^{ 2 }$ so we would just have to substitute the valors on it: $\left[ \frac { 100(100+1) }{ 2 } \right] ^{ 2 }=25502500$

Bonus: to generalise this I used the Telescopic Method. So, in telescopic method, you want to express ${ n }^{ 3 }$ like a subtraction. For example, if we use a couple of transformations in the expression $\left( n+1 \right) ^{ 4 }-{ n }^{ 4 }$ we would get what we want:

$\frac { 1 }{ 4 } \left[ \left( n+1 \right) ^{ 4 }-n^{ 4 }-6n^{ 2 }-4n-1 \right] =n^{ 3 }$

Now if we substitute some valors in the expression we obtained:

$\frac { 1 }{ 4 } \left[ 2^{ 4 }-1^{ 4 }-6\times 1^{ 2 }-4\times 1-1 \right] =1^{ 3 }\\ \frac { 1 }{ 4 } \left[ 3^{ 4 }-2^{ 4 }-6\times 2^{ 2 }-4\times 2-1 \right] =2^{ 3 }\\ \frac { 1 }{ 4 } \left[ 4^{ 4 }-3^{ 4 }-6\times 3^{ 2 }-4\times 3-1 \right] =3^{ 3 }\\ \frac { 1 }{ 4 } \left[ 5^{ 4 }-4^{ 4 }-6\times 4^{ 2 }-4\times 4-1 \right] =4^{ 3 }$

We can observe that if we sum all that up, the first and second terms inside the brackets cancell each other, the only terms which wouldn't be cancelled would be $(n+1)^{ 4 }$ and $-1^{ 4 }$ after that we have a $6$ multiplying a sume of the first n squares, so we substitute the formula or deduce it with this method. Next, there is a $4$ multiplying the sum of the first n numbers also known as "Gauss Sum". So we again substitute the formula or deduce it and at the end we have a $-1$ which is being subtracted n times so we can write it as $-n$ . So the final expression would look like:

$\frac { 1 }{ 4 } \left[ \left( n+1 \right) ^{ 4 }-1^{ 4 }-6\times \left( \frac { n(n+1)(2n+1) }{ 6 } \right) -4\times \left( \frac { n(n+1) }{ 2 } \right) -n \right] =\left[ \frac { n(n+1) }{ 2 } \right] ^{ 2 }$

And that's how we can obtain the general "n" formula. It's pretty curious because it is the square of the Gauss Sum. The title of this problem is in memory of G. H. Hardy, father of divergent sums.