Question

How do you do series with an n

Answer 1

The problem can be solved step by step, if we know certain basic rules of summation.  Following rules assume summation limits are identical.
$\sum{a+b}=\sum{a}+\sum{b}$
$\sum{kx}=k\sum{x}$
$\sum_{r=1}^n{1}=n$
$\sum_{r=1}^n{r}=n(n+1)/2$

Armed with the above rules, we can split up the summation into simple terms:
$\sum_{r=1}^n{40r-21n+8}=n$
$=40\sum_{r=1}^n{r}-21n\sum_{r=1}^n{1}+8\sum_{r=1}^n{1}$
$=40\frac{n(n+1)}{2}-21n^2+8n$
$=20n(n+1)-21n^2+8n$
$=28n-n^2$

=> (a) f(x)=28n-n^2
=> f'(x)=28-2n 
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0  therefore f(14) is a maximum
(b) f(x) is a maximum when n=14

(c) the maximum value of f(x) is f(14)=196