If I have a list of 5 numbers each of them following another [1,2,3,4,5] and I want to multiply all of them together.
How do I write (1*2)+(1*3)+(1*4)+(1*5)+(2*3)+(2*4)+(2*5)+(3*4)+(3*5)+(4*5)
But with 4000 numbers in the list instead of 5 without having to write every instance out?
Is it possible to make a rule and calculate it in a common calculator?
Table:
\(\begin{array}{|r|r|r|r|r|r} \hline \text{multiply} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & \color{red}1 & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline 2 & \color{grey}2 & \color{red}4 & \mathbf{6} & \mathbf{8} & \mathbf{10} \\ \hline 3 & \color{grey}3 & \color{grey}6 & \color{red}9 & \mathbf{12} & \mathbf{15} \\ \hline 4 & \color{grey}4 & \color{grey}8 & \color{grey}12 & \color{red}16 & \mathbf{20} \\ \hline 5 & \color{grey}5 & \color{grey}10 & \color{grey}15 & \color{grey}20 & \color{red}25 \\ \hline sum & \color{blue}15 & \color{blue}30 & \color{blue}45 & \color{blue}60 & \color{blue}75 & \color{blue}\text{arithmetical sequence} \\ \hline \end{array} \)
\(\text{Let the sum of all numbers in the table $\mathbf{S} = \color{blue}{15+30+45+60+75} = \color{blue}225 $} \\ \begin{array}{rcll} \text{Let the sum we are looking for } \mathbf{s=} \\ && (1*2)+(1*3)+(1*4)+(1*5)\\ && +(2*3)+(2*4)+(2*5) \\ && +(3*4)+(3*5) \\ && +(4*5) \\ &=& \mathbf{2+3+4+5} \\ && \mathbf{+6+8+10} \\ && \mathbf{+12+15} \\ && \mathbf{+20} \\ &=& 85\\ \end{array} \\ \begin{array}{rcll} \text{Let the sum of all square numbers in the diagonal of the table $\mathbf{q} =$} \color{red}{1+4+9+16+25}=\color{red}55 \\ \end{array} \)
So we see the formula: \(\boxed{S = 2s + q}\) or \(\boxed{s = \dfrac{1}{2}\cdot (S-q) }\)
\(\mathbf{ S=\ ?}\)
\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)d \quad | \quad d=a_1 \\ &=& a_1+(n-1)a_1 \\ &=& a_1+ na_1-a_1 \\ &=& na_1 \quad | \quad a_1=(1+n) \dfrac{n}{2} \\ &=& (1+n) \dfrac{n^2}{2} \\\\ S &=& (a_1 + a_n)\cdot \dfrac{n}{2} \\ &=& \left((1+n) \dfrac{n}{2} +(1+n) \dfrac{n^2}{2} \right)\cdot \dfrac{n}{2} \\ &=& \Big((1+n) +(1+n) n \Big)\cdot \dfrac{n^2}{4} \\ &=& (n^2+2n+1 )\cdot \dfrac{n^2}{4} \\ &=& (n+1)^2\cdot \dfrac{n^2}{4} \\ \mathbf{S} & \mathbf{=} & \mathbf{\left[\dfrac{(n+1)\cdot n}{2}\right]^2} \\ \hline \end{array}\)
\(\mathbf{ q=\ ?}\)
\(\begin{array}{|rcll|} \hline \mathbf{q} & \mathbf{=} & \mathbf{\dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3}} \quad | \quad \text{from the formula collection} \\ \hline \end{array} \)
\(\mathbf{ s=\ ?}\)
\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{ \dfrac{1}{2}\cdot (S-q) } \\ &=& \dfrac{1}{2}\cdot \left(\left[\dfrac{(n+1)\cdot n}{2}\right]^2 - \dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3} \right) \\ &=& \dfrac{(n+1)\cdot n}{4}\cdot \left[ \dfrac{(n+1)\cdot n}{2} - \dfrac{(2n+1)}{3} \right] \\ &=& \dfrac{(n+1)\cdot n}{24}\cdot \Big[ 3n(n+1) - 2(2n+1) \Big] \\ &=& \dfrac{(n+1)\cdot n\cdot ( 3n^2-n-2)}{24} \\ &=& \dfrac{(n+1)\cdot n\cdot (3n+2)\cdot (n-1)}{24} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(n-1)\cdot n \cdot (n+1)\cdot (3n+2)}{24} } \\ \hline \end{array}\)
Example: \([1,2,3,4,5]\)
So \(n = 5 \)
\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(5-1)\cdot 5 \cdot (5+1)\cdot (3\cdot 5+2)}{24} } \\ &=& \dfrac{4\cdot 5 \cdot 6\cdot 17}{24} \\ &=& 5 \cdot 17 \\ &\mathbf{=}& \mathbf{85}\ \checkmark\\ \hline \end{array}\)
Example:\( [1,2,3,4,5, \ldots ,4000]\)
So \(n = 4000\)
\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(4000-1)\cdot 4000 \cdot (4000+1)\cdot (3\cdot 4000+2)}{24} } \\ &=& \dfrac{3999\cdot 4000 \cdot 4001\cdot 12002 }{24} \\ &=& 1333\cdot 1000 \cdot 4001\cdot 6001 \\ &\mathbf{=}& \mathbf{32\ 005\ 331\ 333\ 000} \\ \hline \end{array} \)
