PLEASE HELP!!!!!!!!!!
1 answer:
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Here's a pattern to consider:
1+100=101
2+99=101
3+98=101
4+97=101
5+96=101
.....
This question relates to the discovery of Gauss, a mathematician. He found out that if you split 100 from 1-50 and 51-100, you could add them from each end to get a sum of 101. As there are 50 sets of addition, then the total is 50×101=5050
So, the sum of the first 100 positive integers is 5050.
Quick note
We can use a formula to find out the sum of an arithmetic series:
Where s is the sum of the series and n is the number of terms in the series. It works for the above problem.
1+100=101
2+99=101
3+98=101
4+97=101
5+96=101
.....
This question relates to the discovery of Gauss, a mathematician. He found out that if you split 100 from 1-50 and 51-100, you could add them from each end to get a sum of 101. As there are 50 sets of addition, then the total is 50×101=5050
So, the sum of the first 100 positive integers is 5050.
Quick note
We can use a formula to find out the sum of an arithmetic series:
Where s is the sum of the series and n is the number of terms in the series. It works for the above problem.
Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions </u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
(a)
First, we take its derivative
Solve N'(t)=0
Simplifying
Solving for t
Only t=1 belongs to the valid interval
Taking the second derivative
Which is always positive, so t=1 is a minimum
(b)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
N(15)=4391.7 is a maximum