The K-Square and the first n positive integers

How to derive the summation formula

\ (1^2+2^2+3^2+\cdots+n^2=\frac{n (n+1) (2n+1)}{6} \)

And

\ (1^3+2^3+3^3+\cdots+n^3=\frac{n^2 (n+1) ^2}{4} \)

It? This requires a bit of skill.

First look at an identity:

\ ((n+1) ^3-n^3 = 3n^2+3n+1 \)

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

On the summation, \ ((n+1) ^3-n^3 \) will have only two entries (the middle of the elimination), \ (3n+1 \) is arithmetic progression, the summation formula is known. In this way, we can find the summation formula of two times. In the same way, you can introduce a three-square summation formula:

\ ((n+1) ^4-n^4 = 4n^3 + 6n^2 + 4n + 1 \)

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

The sum of the n+1, \ (() ^4-n^4 \) will have only two entries, \ (6n^2 \) and \ (4n+1 \) Summation formula is known (the former can be deduced from the above two-square summation formula, the latter is arithmetic progression. This will enable the introduction of a three-time formula for summation. All the time, the summation formula can be obtained by pushing it all the way down.

The K-Square and the first n positive integers