Question

Find the sum of all multiples of 3 between 1 and 1000

Answer 1

what is the first multiple of 3 between 1 and 1000 ?

Answer 2

Um its 3

Answer 3

and last ?

Answer 4

999*

Answer 5

so, can you tell me how many multiples of 3 are there in total ?

Answer 6

yes, last is 999

Answer 7

Is it 333 terms?

Answer 8

yes! thats correct!

Answer 9

now do you know the formula for sum ?

Answer 10

All the multiples of three between 1 and 1000 and may be written in the form \(3n\) where \(1\le n\le333\) (which gives a lowest number \(3\) and highest number \(999\)). Note, then, that if we denote our first multiple \(a_1=3\), we may define a sequence such that the \(n\)-th multiple is given by \(a_n=a_1+3(n-1)\). Do you recognize this? It's an arithmetic progression.

The sum of all these multiples, then, corresponds to an arithmetic series of all 333 of our terms. Recall that for an arithmetic series the \(n\)-th sum is given by \(S_n=\frac12n(a_1+a_n)\). Thus our sum is \(S_{333}=\frac12(333)(3+999)=\frac12(333)(1002)=501(333)=166833\).

Answer 11

Yes, I do

Answer 12

great! just plug in values :)

Answer 13

Oh i got it! Thank you!

Answer 14

Thanks to you too @oldrin.bataku ^_^

Answer 15

@oldrin.bataku your answer was good, but try to engage the user next time :)

Answer 16

The general formula is derived as follows for an arithmetic series with difference \(k\) and initial term \(a_1\). Let \(S_n\) be our sum for the first \(n\) terms so$$S_n=a_1+(a_1+k)+(a_1+2k)+\cdots+(a_1+(n-3)k)+(a_1+(n-2)k)+(a_1+(n-1)k)$$ Now, rewrite this sum in backwards order$$S_n=(a_1+(n-1)k)+(a_1+(n-2)k)+(a_1+(n-3)k)+\cdots+(a_1+2k)+(a_1+k)+a_1$$Adding these two sums term by term we find$$\begin{align*}S_n+S_n&=(a_1+a_1+(n-1)k)+(a_1+a_1+(n-1)k)+\cdots+(a_1+a_1+(n-1)k))+(a_1+a_1+(n-1)k))\\&=n(a_1+a_1+(n-1)k)\\&=n(a_1+a_n)\end{align*}$$In total, on the right hand side, we had \(n\) terms of \(a_1+a_n\) so we condensed it to \(n(2a_1+(n-1)k)\). Recognizing that \(a_1+(n-1)k\) merely determines our \(n\)-th term we replace it with \(a_n\).

On the left, however, we recognize that \(S_n+S_n=2S_n\)$$2S_n=n(a_1+a_n)$$Solving for \(S_n\), we have \(S_n=\frac12n(a_1+a_n)\)/

Answer 17

Oops. I meant we had \(n\) terms of \(a_1+a_1+(n-1)k\) so we condensed into \(n(a_1+a_1+(n-1)k)\) and then substituted \(a_1+(n-1)k=a_n\) to get into the familiar form \(n(a_1+a_n).\)