The lowest one is 4, just multiply the first fraction by 2
No they're not proportional, and it's four*.
4/2 = 2
20/6 = 3.333333333334 or 3 1/3
The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.




Armed with the above rules, we can split up the summation into simple terms:





=> (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