Question 1162191: Solve for the sum of all positive integers, from 5 to 1255, that can be divided by 5.
a. 158,130
b. 240,840
c. 76,210
d. 99,040
Found 2 solutions by MathLover1, jim_thompson5910:
Answer by MathLover1(20850)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
We're adding all the multiples of 5 from 5 on up to 1255
We see that,
5 = 5*1
10 = 5*2
15 = 5*3
and so on, until,
1255 = 5*251 (note how 1255/5 = 251)
This shows we have 251 multiples of 5 being added up like so
5*1 + 5*2 + 5*3 + ... + 5*251
In which we can factor out the common 5's for each term (distributive property)
5*1 + 5*2 + 5*3 + ... + 5*251 = 5(1+2+3+...251)
The question is now: "What is the sum of the first 251 positive integers?" or "what is the sum of the positive integers from 1 to 251?"
The formula to answer this is
S = n*(n+1)/2
which finds the sum of integers from 1 to n
So,
1+2+3+...+n = n*(n+1)/2
1+2+3+...+251 = 251*(251+1)/2
1+2+3+...+251 = 251*126
1+2+3+...+251 = 31626
and,
5*1 + 5*2 + 5*3 + ... + 5*251 = 5(1+2+3+...251)
5*1 + 5*2 + 5*3 + ... + 5*251 = 5(31626)
5*1 + 5*2 + 5*3 + ... + 5*251 = 158130
You can use a free online tool such as Wolfram Alpha to help check we have the right answer
The input "Sum[5*n,n=1..251]" means we are summing terms of the form 5*n from n = 1 to n = 251. The fancy/weird looking "E" represents the greek letter sigma, to indicate a summation.
Another way to check is to add up each individual item in the series; though that is a tedious task which will take a lot of time. Not to mention there is a fairly high chance of user error. Luckily there are various programming and scripting languages to make the task much easier.
Answer: 158130 (choice A)
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