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slavikrds [6]
3 years ago
6

How do you do series with an n

Mathematics
1 answer:
SOVA2 [1]3 years ago
4 0
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
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