This question was uploaded on 22/01/22 on social media accounts.
Solution:
From the given equations:
\[\color{blue} {a_1=0}\]
\[\color{blue} {a_2=a_1+2(1)}\]
\[\color{blue} {a_3=a_2+2(2)}\]
\[\color{blue} {a_4=a_3+2(3)}\]
.
.
.
\[\color{blue} {a_n=a_{n-1}+2(n-1)}\]
By adding these all equation:
\[\color{red} {a_n=2\frac{(n-1)(n)}2=(n-1)(n)}\]
Now because of the general equation, we can calculate the value of the multiplication:
\[\color{black} {\left(\frac{a_3}{a_4}\right)\left(\frac{a_5}{a_6}\right)\left(\frac{a_7}{a_8}\right).....\left(\frac{a_{4043}}{a_{4044}}\right)}\]
\[=\color{black} {\left(\frac{(2)(3)}{(3)(4)}\right)\left(\frac{(4)(5)}{(5)(6)}\right)\left(\frac{(6)(7)}{(7)(8)}\right).....\left(\frac{(4042)(4043)}{(4043)(4044)}\right)}\]
\[=\frac{2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot4042\cdot4043}{3\cdot4\cdot5\cdot6\cdot\cdot\cdot\cdot4043\cdot4044}\]
\[=\frac{2}{4044}=\frac1{2022}\]
