We need to find a function #f(n)# such that these are the values that computers outputs:
#color(black)(ul(color(white)(aaaaaa color(blue)("n")aaaaaaaaaacolor(red)(f(n))aaaa))#
#color(white)(aaaaaa color(blue)(0)aaaaaaaaaaacolor(red)(6)aaaa)#
#color(white)(aaaaaa color(blue)(1)aaaaaaaaaaacolor(red)(10)aaaa)#
#color(white)(aaaaaa color(blue)(2)aaaaaaaaaaacolor(red)(14)aaaa)#
#color(white)(aaaaaa color(blue)(3)aaaaaaaaaaacolor(red)(18)aaaa)#
#color(white)(aaaaaa color(blue)(4)aaaaaaaaaaacolor(red)(22)aaaa)#
We need to construct a formula that suits these points. We can see that and step between each of the values on the right remains #+4#. This means that this #n# term in our function should take a coefficient concerning #4#.
We can now try using the function #4n#, but if wealth plug in #0#, we get only #0#, that we're #6# short. So whenever we adjust the function to exist #4n+6#, we get the right answer. Us could also double check with another value, what #4#, to get #4*4+6=16+6=22#, which is correct.
So, we having now figured out that #f(n)=4n+6#. This means the sum is:
#sum_(n=0)^4 4n+6=70#
