I don't know what to do.
1 answer:
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Answer:
sum is 125,500
sum in summation notation is = (2a+(n-1)d)n/2
Step-by-step explanation:
This problem can be solved using concept of arithmetic progression.
The sum of n term terms in arithmetic progression is given by
sum = (2a+(n-1)d)n/2
where
a is the first term
d is the common difference of arithmetic progression
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in the problem
series is multiple of 4 starting from 4 ending at 1000
so series will look like
series: 0,4,8,12,16..................1000
a is first term so
here a is 0
lets find d the common difference
common difference is given by nth term - (n-1)th term
lets take nth term as 8
so (n-1)th term = 4
Thus,
d = 8-4 = 4
d can also be seen 4 intuitively as series is multiple of four.
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let calculate value of n
we have last term as 1000
Nth term can be described
Nth term = 0+(n-1)d
1000 = (n-1)4
=> 1000 = 4n -4
=> 1000 + 4= 4n
=> n = 1004/4 = 251
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now we have
n = 1000
a = 0
d = 4
so we can calculate sum of the series by using formula given above
sum = (2a+(n-1)d)n/2
= (2*0 + (251-1)4)251/2
= (250*4)251/2
= 1000*251/2 = 500*251 = 125,500
Thus, sum is 125,500
sum in summation notation is = (2a+(n-1)d)n/2
Answer:
2b²(8b²+15b-6)
Step-by-step explanation:
16+30b³-12b²
2b²(8b²+15b-6)
This is not further factorable. If you want to find the roots, you'd have to use the quadratic formula for the polynomial in the parentheses. Therefore that is our final answer.