Encontrar la suma de la serie a continuación

Question

Encontrar la suma

$$(1\cdot2)+(1\cdot3)+(1\cdot4)+\cdots+(1\cdot2015)+(2\cdot3)+(2\cdot4)+\cdots+(2\cdot2015)+\cdots+(2014\cdot2015)$$

Lo he probado...

Estamos buscando para el $$S(n)=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij$$ when $n = 2015$ $$S(n)=\sum_{1\le i<j\le n}ij$$ $% $ $=\frac 12\left((\sum_{i=1}^n i)^2-\sum_{i=1}^ni^2\right)$

$$S(n)=\frac 12\left(\frac{n^2(n+1)^2}4-\frac{n(n+1)(2n+1)}6\right)$$ $$=\frac{(n-1)n(n+1)(3n+2)}{24}$$

Y así $$\boxed{S(2015)=\frac{2014\times 2015\times 2016\times 6047}{24}}$ $

¿Es la solución equivocada o no?

Answer 1

Sumando a lo largo de las diagonales que se le da un buen simétrica de la derivación de la solución, utilizando los coeficientes binomiales y minimizar desordenado manipulación de fracciones. $$\begin{align} S(n)&=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij \color{lightgray}{=\sum_{1\le i<j\le n} ij}\\\\ &=\begin{cases}\begin{matrix} \color{orange}{1\cdot 2} &+1\cdot 3 &+1\cdot 4 &+\cdots &+\color{purple}{1\cdot (n-2)} &\color{blue}{+1\cdot (n-1)} &\color{green}{+1\cdot n}&\\ &\color{orange}{+2\cdot 3} &+2\cdot 4 &+\cdots &+2\cdot (n-2) &\color{purple}{+2\cdot (n-1)} &\color{blue}{+2\cdot n}&\\ & &\color{orange}{+3\cdot 4} &+\cdots &+3\cdot (n-2) &+3\cdot (n-1) &\color{purple}{+3\cdot n}& \\ & & &\ddots & & &\vdots &\\ & & & & & &\color{orange}{+(n-1)\cdot n}& \end{de la matriz}\end{casos}\\\\ &\color{gris claro}{=\sum_{i=1}^{n-1}\sum_{i=1}^{n-r}i(i+r) \qquad\qquad\text{el uso de}\ r=j-i}\\ &\color{gris claro}{=\sum_{m=1}^{n-1}\sum_{i=1}^mi(n-m+i) \qquad\text{el uso de}\ m=n-r}\\\\ y=\begin{cases}\begin{array}{r} \color{green}{1\cdot n}&\\ \color{blue}{+1\cdot (n-1)+2\cdot n}&\\ \color{purple}{+1\cdot (n-2)+2\cdot (n-1)+3\cdot n}&\\ \vdots\\ \color{orange}{+1\cdot 2+2\cdot3 +3\cdot 4+\cdots +(n-3)\cdot (n-2)+(n-2)\cdot (n-1)+(n-1)\cdot n}\end{array}\end{casos}\\\\ &\color{gris claro}{=\sum_{m=1}^{n-1}\sum_{i=1}^m ni-i(m-i)}\\ &\color{gris claro}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{m=1}^m\sum_{k=i+1}^m i\right]}\\ &\color{gris claro}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{k=2}^m\sum_{i=1}^{k-1}i\right]}\\ &\color{gris claro}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{k=2}^m\binom k2\right]}\\ &\color{gris claro}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\binom {m+1}3\right]}\\ y=\begin{cases}\begin{array}{l}\quad \color{Green}{n}&\\ \color{blue}{+3n-1}\\ \color{purple}{+6n-4}&\\ +10n-10&\\ \vdots\\ +\binom {m+1}2n-\binom {m+1}3&\\ \vdots \\ \color{orange}{+\binom n2 n-\binom n3}\end{array}\end{casos}\\\\ &=\sum_{m=1}^{n-1}\binom {m+1}2 n-\binom {m+1}3\\\\ &=\binom {n+1}3n-\binom {n+1}4\\\\ &=n\binom {n+1}3-\frac{n-2}4\binom{n+1}3\\\\ &=\frac{3n+2}4\binom{n+1}3\qquad\blacksquare\\\\ S(2015)&=\frac {6047}4\binom{2016}3=2\;061\;359\;653\;080\qquad\blacksquare \end{align}$$

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Nota

La suma también puede ser calculada mediante la suma de las columnas en lugar de filas (es decir, el intercambio de $i$$j$) como sigue:

$$\begin{align} S(n) &=\sum_{i=1}^{n-1}\sum_{j=i+1}^n ij\\ &=\sum_{1\le i<j\le n}ij\\ &=\sum_{j=2}^n\sum_{i=1}^{j-1}ij\\ &=\sum_{j=1}^n j\sum_{i=1}^{j-1}\binom i1\\ &=\sum_{j=1}^n j\binom j2\\ &=\sum_{j=1}^n (j+1)\binom j2-\binom j2\\ &=\sum_{j=1}^n 3\binom {j+1}3-\binom j2\\ &=3\binom {n+2}4-\binom {n+1}3\\ &=\frac {3(n+2)}4\binom {n+1}3-\binom {n+1}3\\ &=\frac {3n+2}4\binom {n+1}3\qquad\blacksquare \end{align}$$