We need to find a function #f(n)# such that these are the values that it 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 fits these points. We can see that the step between each of the values on the right is #+4#. This means that the #n# term in our function should have a coefficient of #4#.
We can now try using the function #4n#, but if we plug in #0#, we get only #0#, so we're #6# short. So if we adjust the function to be #4n+6#, we get the right answer. We can also double check with another value, say #4#, to get #4*4+6=16+6=22#, which is correct.
So, we have now figured out that #f(n)=4n+6#. This means the sum is:
#sum_(n=0)^4 4n+6=70#
