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# C1 Arithmitic series help Watch

1. Merry Christmas everyone,

I'm just a little confused on some questions that frequently pop up in the papers.

Thay say something like,

Prove that the sum of the first n positive integers is given by
1/2 n(n + 1)

And I know that you show something like

Sn= 1+2+3+...+n
Sn= n+(n-1)+(n-2)+...+1
2Sn= n(n+1)
Sn= 1/2 n(n+1)

But it makes no sense at all :s

Why does this happen and why does it become to that formula?
2. Sn = n + (n-1) + (n-2) + ... + 1
Rearrange the terms into pairs that each total n+1:
Sn = [n + 1] + [(n-1) + 2] + [(n-2) + 3] + ...
There will always be n/2 such pairs in the sequence, so:
Sn = n/2(n+1)
3. S_n = 1 + 2 + ... + n should be obvious

S_n = n + (n-1) + (n-2) + ... + 1 is S_n written backwards (e.g. 1 + 2 + 3 + 4 + 5 = 5 + 4 + 3 + 2 + 1, they're the same thing)

Now if you add the first line and the second line, on the left hand side you get S_n + S_n = 2S_n.

On the right hand side, add the terms like they're in a column.

So first add the 1 and the n, to get 1 + n
Then add the 2 + (n-1), to get 1 + n
Then the 3 + (n-2), to get 1 + n
...
Then the n + 1, to get 1 + n

You get 1 + n every time and there are n terms, so all together you have n(1+ n) on the right hand side.

Left hand side = 2S_n
Right hand side = n(n+1)

So 2S_n = n(n+1)
4. (Original post by Swayum)
S_n = 1 + 2 + ... + n should be obvious

S_n = n + (n-1) + (n-2) + ... + 1 is S_n written backwards (e.g. 1 + 2 + 3 + 4 + 5 = 5 + 4 + 3 + 2 + 1, they're the same thing)

Now if you add the first line and the second line, on the left hand side you get S_n + S_n = 2S_n.

On the right hand side, add the terms like they're in a column.

So first add the 1 and the n, to get 1 + n
Then add the 2 + (n-1), to get 1 + n
Then the 3 + (n-2), to get 1 + n
...
Then the n + 1, to get 1 + n

You get 1 + n every time and there are n terms, so all together you have n(1+ n) on the right hand side.

Left hand side = 2S_n
Right hand side = n(n+1)

So 2S_n = n(n+1)
Wait, if n = 1, why does n - 1 = 2?

Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

5. (Original post by im so academic)
Wait, if n = 1, why does n - 1 = 2?

Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

Given a=1 in this case, and d=1 we can simplify your formula to the above.
Given a=1 in this case, and d=1 we can simplify your formula to the above.
Oh, thank you very much!

But is it possible to do it the "[2a + (n-1)d]" way and then simplify it, right?
7. (Original post by im so academic)
Wait, if n = 1, why does n - 1 = 2?

Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

If n = 1, you're asking to sum the series with just one term: 1. So there doesn't exist an (n-1)th term in this context. The question is to sum

1 + 2 + 3 + ... + n

You can let n = 3, in which case you're summing 1 + 2 + 3 = 6. The way our method works is that

S_3 = 1 + 2 + 3 (forwards)
S_3 = 3 + 2 + 1 (backwards)

Add those lines to get 2S_3 = (1+3) + (2+2) + (3+1) = 4 + 4 + 4 = 3*4 = 12
So S_3 = 6

You can let n = 1, in which case you're summing 1 = 1. The way our methods works is that

S_1 = 1 (forwards)
S_1 = 1 (backwards)

Add and you get 2S_1 = 1 + 1 = 2

So S_1 = 1

S_n = 0.5n(2a + d(n-1)) is the sum starting from the number a and with any difference between two terms. The person above asked for the formula starting from the number 1 and with a difference of 1 between any two numbers.

Clearly, starting at a with a difference of d is a more general way of doing it, and that formula is derived in exactly the same way as writing the series backwards and adding it.
8. (Original post by im so academic)
Oh, thank you very much!

But is it possible to do it the "[2a + (n-1)d]" way and then simplify it, right?
Yes, just scribbled it down.

Sn=n/2[2a+d(n-1)]

We know a=1 and d=1

Sn=n/2[(2x1)+1(n-1)]

Sn=n/2[(2+n-1)]

Sn=n/2(n+1)

You could then multiply both sides by two and then divide again, but its trivial. And that's the 'proof'. Enjoy
9. (Original post by im so academic)
Wait, if n = 1, why does n - 1 = 2?

Also, shouldn't the formula be Sn = n/2[2a + (n-1)d]

Oh I see what you've misunderstood. I said that

"S_n = 1 + 2 + ... + n should be obvious

S_n = n + (n-1) + (n-2) + ... + 1 is S_n written backwards"

We're not saying that the terms that are above and below each other are EQUAL. No no no. We're saying, let's add them TOGETHER. It's like

1 + 2 + 3 + 4 is still the same as 4 + 3 + 2 + 1 but written backwards, isn't it? The quantity doesn't change.

But the point is to now add the first terms together: 1 and 4

Then add second terms together: 2 and 3

Then third terms: 3 and 2

Then fourth terms: 4 and 1

Notice all of these are 5! 1 + 4 = 5, 2 + 3 = 5, 3 + 2 = 5, 4 + 1 = 5. That's the point - they all equal the same thing, and there are n of them, so we can add them together easily.
Yes, just scribbled it down.

Sn=n/2[2a+d(n-1)]

We know a=1 and d=1

Sn=n/2[(2x1)+1(n-1)]

Sn=n/2[(2+n-1)]

Sn=n/2(n+1)

You could then multiply both sides by two and then divide again, but its trivial. And that's the 'proof'. Enjoy
In this question, they wouldn't accept that as a proof because you haven't explained why S_n = 0.5n(2a + d(n-1)). Usually, you can just quote that formula in C1 questions, but you wouldn't be allowed to/it wouldn't be acceptable in this question.
11. (Original post by Swayum)
In this question, they wouldn't accept that as a proof because you haven't explained why S_n = 0.5n(2a + d(n-1)). Usually, you can just quote that formula in C1 questions, but you wouldn't be allowed to/it wouldn't be acceptable in this question.
Yes, fair enough, I see that, but academic was asking why the other formula couldn't be used I think. Hopefully I don't get this question in January, hopefully something more exciting...

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