A Long Equation III

Chew-Seong Cheong’s solution is good, but I think there’s an interesting approach as follows, $\begin{aligned} \because 1\times 2 &=\dfrac{1}{3}\times \left( 1\times 2\times 3-0\times 1\times 2 \right) \\ 2\times 3 &=\dfrac{1}{3}\times \left(2\times 3\times 4 - 1 \times 2 \times 3 \right) \\ 3\times 4 &=\dfrac{1}{3}\times \left(3\times 4\times 5-2\times 3\times 4\right) \\ \end{aligned}$ $\therefore 1\times 2+2\times 3+3\times 4=\dfrac{1}{3}\times 3\times 4\times 5=20$ $\cdots \cdots \cdots \cdots$ $\begin{aligned} S &=\dfrac{1}{3}\times (1\times 2\times 3-0\times 1\times 2)+\dfrac{1}{3}\times (2\times 3\times 4-1\times 2\times 3)+\cdots +\dfrac{1}{3}\times (100\times 101\times 102-99\times 100\times 101) \\ &=\dfrac{1}{3}\times (100\times 101\times 102-0) \\ &=343400 \end{aligned}$

$\begin{aligned} S_n & = 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + n(n+1) \\ & = \sum_{k=1}^{100} k(k+1) \\ & = \sum_{k=1}^{100} k^2 + \sum_{k=1}^{100} k \\ & = \frac {n(n+1)(2n+1)}6 + \frac {n(n+1)}2 \\ & = \frac {n(n+1)(2n+1+3)}6 \\ & = \frac {n(n+1)(n+2)}3 \end{aligned}$

Therefore $S_{100} = \dfrac {100(101)(102)}3 = \boxed {343400}$ .

This is a direct derivation of Chew-Seong Cheong's formula from first principles. I learned how to do this by myself and on my own to eliminate busywork when I was about the author's age. They are called finite forward differences:

$\begin{array}{|c|l|} \hline 2,8,20,40,70,112 & \text{original function values} \\ \hline 6,12,20,30,42 & \text{first differences} \\ \hline 6,8,10,12 & \text{second differences} \\ \hline 2,2,2 & \text{third differences} \\ \hline \end{array}$

Here are the finite difference polynomials:

$\begin{array}{rl} 0 & 1 \\ 1 & n-1 \\ 2 & \frac{1}{2} (n-2) (n-1) \\ 3 & \frac{1}{6} (n-3) (n-2) (n-1) \\ 4 & \frac{1}{24} (n-4) (n-3) (n-2) (n-1) \\ 5 & \frac{1}{120} (n-5) (n-4) (n-3) (n-2) (n-1) \\ 6 & \frac{1}{720} (n-6) (n-5) (n-4) (n-3) (n-2) (n-1) \\ 7 & \frac{(n-7) (n-6) (n-5) (n-4) (n-3) (n-2) (n-1)}{5040} \\ 8 & \frac{(n-8) (n-7) (n-6) (n-5) (n-4) (n-3) (n-2) (n-1)}{40320} \\ 9 & \frac{(n-9) (n-8) (n-7) (n-6) (n-5) (n-4) (n-3) (n-2) (n-1)}{362880} \\ 10 & \frac{(n-10) (n-9) (n-8) (n-7) (n-6) (n-5) (n-4) (n-3) (n-2) (n-1)}{3628800} \\ \end{array}$

Using the differences down the left edge of the differences triangle:

$\bm{2}(1)+\bm{6}(n-1)+\bm{6}(\frac12(n-2)(n-1))+\bm{2}(\frac16(n-3)(n-2)(n-1))$

Which simplifies to:

$\frac{1}{3} n (n+1) (n+2)$

Evaluating that expression at $n=100$ gives $\frac{100\times 101\times 102}{3} \Rightarrow 343400$ .

This procedure, written as a Wolfram Mathematica expression, is:

$\text{generator}=\text{Function}[\{\text{diffs},\text{start},\text{step}\}, \\ \text{FullSimplify}\left[\text{Evaluate}\left[\text{Expand}\left[\sum _{i=0}^{\text{Length}[\text{diffs}]-1} \frac{\text{diffs}[[i+1]] \prod _{j=0}^{i-1} \left(\frac{\text{\#1}-\text{start}}{\text{step}}-j\right)}{i!}\right]\right]\right]\& ];$

1×2+2×3+3×4+......+99×100+100×101 =2(1+3)+4(3+5)+6(5+7)+......+100(99+101) =2×4+2×2×2×4+2×3×3×4+......+2×50×4×50 =8+8×2²+8×3²+......+8×50² =8(1+2²+3²+......+50²) =8×42925 =343400

Using identities： (n+1)³=n³+3n²+3n+1 → (n+1)³-n³=3n²+3n+1

n³-(n-1)³=3(n-1)²+3(n-1)+1 ......

3³-2³=3×(2²)+3×2+1

2³-1³=3×(1²)+3×1+1.

If I add these n sides of this equation, To:(n+1)³-1=3(1²+2²+3²+....+n²)+3(1+2+3+...+n)+n,

∵1+2+3+...+n=(n+1)n/2

∴n³+3n²+3n=3(1²+2²+3²+....+n²)+3(n+1)n/2+n

∴ 1²+2²+3²+....+n²=n(n+1)(2n+1)/6

then 50×51×101÷6=42925