Challenge Level

In 1654 Blaise Pascal published a general method for summing powers of positive integers, i.e. summing all the series. $$S_1 = 1 + 2 + 3 + ... + n$$ $$S_2 = 1^2 + 2^2 + 3^2 + ... + n^2$$ $$S_3 = 1^3 + 2^3 + 3^3 + ... + n^3$$ $$\dots$$ $$S_r = 1^r + 2^r + 3^r + ... + n^r$$Pascal's method uses the coefficients which appear in Pascal's triangle and in the Binomial Theorem,
first finding $S_1$, and then using $S_1$ to find $S_2$, and then using both to find $S_3$, and so on. The method applies, where $r$ is any fixed positive integer, to: $$S_r =\sum_{k=1}^n k^r.$$

Case 1: $r = 1$

Simplify $(k+1)^2 - k^2$ and hence show that: $$ \sum_{k=1}^n [(k+1)^2-k^2] = {2\choose 1}S_1 +n. $$ Write the sum out in full for $n = 6$. Use this method to prove that $S_1=n(n+1)/2$.

Case 2: $r = 2$

Simplify $(k+1)^3 - k^3$ and hence show that $$ \sum_{k=1}^n[(k+1)^3-k^3] = (n+1)^3-1 = {3\choose 1}S_2 + {3\choose 2}S_1 + n. $$ Hence prove that $S_2 = n(n+1)(2n + 1)/6$.

Case 3: $r = 3$

Use this technique to find $1^3 + 2^3 + 3^3 + ... + 10^3$.

General case

Show that $$ (n+1)^r - (n+1) = {r\choose 1}S_1 + {r\choose 2}S_2 + {r\choose 3}S_3 + \dots {r\choose r-1}S_{r-1}. $$