Respuesta :
Answer:
sum is 125,500
sum in summation notation is [tex]\sum\limits_{n=0}^n a+nd[/tex]= (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 [tex]\sum\limits_{n=0}^n a+nd[/tex]= (2a+(n-1)d)n/2
