some applications of finite calculus

The sum of any number of terms of a series of values of U_x is equal to the difference between two values of another function V_x. To find the sum \sum U_x, find the function V_x such that \Delta V_x=U_x.


    \[1+2+3+\dots+N=\sum_{x=1}^N \fp{x}{1}=\frac{\fp{x}{2}}{2}\Big|_1^{N+1}=\frac{N(N+1)}{2}\]


S=1\cdot 2+2\cdot 3+3\cdot 4+\dots+n(n+1)
U_x=x(x+1)=\fp{(1+x)}{2}=\fp{(a+bx)}{m}\text{ with a=1, b=1, m=2.}

    \begin{align*} \sum_1^n U_x&=\sum_1^n \fp{(1+x)}{2}=\sum \fp{(1+x)}{2} \Big|_1^{n+1}=\frac{\fp{(1+x)}{3}}{3}\Big|_1^{n+1} \\ &=\frac{\fp{(n+2)}{3}}{3}-\frac{\fp{2}{3}}{3}=\fr{3}(n+2)(n+1)n \end{align*}


3 pictures of \sum n(n+1). The first is (I believe) my own creation.

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    \begin{gather*} 3\cdot 1\cdot 2+3\cdot 2\cdot 3+3\cdot 3\cdot 4+3\cdot 4\cdot 5+\dots+3\cdot 13\cdot 14=13\cdot 14\cdot 15 \\ \sum n(n+1)=\frac{n(n+1)(n+2)}{3} \end{gather*}


12p23-17


12p34


Evaluate S=1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\dots +14\cdot 15\cdot 16

    \[S=\sum_{x=3}^{16} x(x-1)(x-2)=\sum_{x=3}^{16} \fp{x}{3}=\frac{\fp{x}{4}}{4}\Big|_3^{17}\]

    \[=\frac{17\cdot 16\cdot 15\cdot 14 - 3\cdot 2\cdot 1\cdot 0}{4}=14,280\]

123p234-18


Evaluate S=2\cdot 5 \cdot 8+5 \cdot 8\cdot 11+8\cdot 11\cdot 14+\dots+20\cdot 23\cdot 26

Using \sum \fp{(a+bx)}{n}=\dfrac{\fp{(a+bx)}{n+1}}{b(n+1)}, with n=3, a=-1, b=3:

    \begin{align*} \sum_{x=3}^9 \fp{(-1+3x)}{3}&=\frac{\fp{(-1+3x)}{4}}{3\cdot 4}\Big|_3^{10}\\ &=\frac{29\cdot 26\cdot 23\cdot 20-8\cdot 5\cdot 2\cdot (-1)}{12} \\ &=28,910 \end{align*}


S=1\cdot 2\cdot 3+3\cdot 4\cdot 5+5\cdot 6\cdot 7+\dots+(2n-1)(2n)(2n+1)

    \begin{align*} &x&&0&&1&&2&&3&&4&&5&\\ &f(x) &&0 &&6 &&66 &&276 &&780 &&1770 &\\ &\Delta f(x) &&6 &&60 &&210 &&504 &&990 && &\\ &\Delta^2 f(x) &&54 &&150 &&294 &&486 && && &\\ &\Delta^3 f(x) &&96 &&144 &&192 && && && &\\ &\Delta^4 f(x) &&48 &&48 && && && && &\\ &\Delta^5 f(x) &&0 && && && && && & \end{align*}

    \begin{align*} V_x&=f(0)+\Delta f(0)x+\Delta^2 f(0) \frac{\fp{x}{2}}{2!}+\Delta^3 f(0) \frac{\fp{x}{3}}{3!}+\Delta^4 f(0) \frac{\fp{x}{4}}{4!}\\ &=0+6x+54\frac{\fp{x}{2}}{2}+96\frac{\fp{x}{3}}{6}+48\frac{\fp{x}{4}}{24}\\ &=6x+27x(x-1)+16x(x-1)(x-2)+2x(x-1)(x-2)(x-3)\\ &=2x^4+4x^3+x^2-x\\ &=2x^3(x+2)+x(x-1) \end{align*}


S=1\cdot 2\cdot 3+4\cdot 5\cdot 6+7\cdot 8\cdot 9+\dots+(3n-2)(3n-1)3n

    \begin{align*} &x&&0&&1&&2&&3&&4&&5&\\ &f(x) &&0 &&6 &&126 &&630 &&1950 &&4680 &\\ &\Delta f(x) &&6 &&120 &&504 &&1320 &&2730 && &\\ &\Delta^2 f(x) &&114 &&384 &&816 &&1410 && && &\\ &\Delta^3 f(x) &&270 &&432 &&594 && && && &\\ &\Delta^4 f(x) &&162 &&162 && && && && &\\ &\Delta^5 f(x) &&0 && && && && && & \end{align*}

    \begin{align*} V_x&=f(0)+\Delta f(0)x+\Delta^2 f(0) \frac{\fp{x}{2}}{2!}+\Delta^3 f(0) \frac{\fp{x}{3}}{3!}+\Delta^4 f(0) \frac{\fp{x}{4}}{4!}\\ &=0+6x+114\frac{\fp{x}{2}}{2}+270\frac{\fp{x}{3}}{6}+162\frac{\fp{x}{4}}{24}\\ &=6x+57x(x-1)+45x(x-1)(x-2)+\frac{27}{4}x(x-1)(x-2)(x-3)\\ &=6x+57x^2-57x+45(x^3-3x^2+2x)+\frac{27}{4}(x^4-6x^3+11x^2-6x)\\ &=\frac{27x^4+18x^3-15x^2-6x}{4}\\ &=\frac{3x}{4}(9x^3+6x^2-5x-2) \end{align*}


Q. What is the general formula for this type of sum,
1\cdot 2\cdot 3+(1+a)(2+a)(3+a)+(1+2a)(2+2a)(3+2a)+\dots
+(1+a(n-1))(2+a(n-1))(3+a(n-1)) ?

(1+a(n-1))(2+a(n-1))(3+a(n-1))
\qquad =a^3n^3+3a^2n^2(2-a)+an(11-12a+3a^2)-(a-1)(a-2)(a-3)

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