Question

Calculate the sum of the multiples of 4 from 0 to 1000

Answer 1

Answer:

sum is 125,500

sum in summation notation is $\sum\limits_{n=0}^n a+nd$= (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 $\sum\limits_{n=0}^n a+nd$= (2a+(n-1)d)n/2