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## Main Question or Discussion Point

i am given an formula Sn= n/2[2a+(n-1)d] and i am told to verify the formula represents the sum of n terms of an arithmetic series. How do i verify this?

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i am given an formula Sn= n/2[2a+(n-1)d] and i am told to verify the formula represents the sum of n terms of an arithmetic series. How do i verify this?

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matt grime

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1. work out a constructive proof (if it's any help i believe there's a famous example of this from a child in class one day, asked to count the number of birds if there war 1 on the first step, 2 on the second, 3 on the third etc). hint, let S(n) be the sum of the first n terms of the series

S(n)=a(1)+a(2)+...+a(n)

well, S(n) also equals

a(n)+...+a(2)+a(1)

and what is a(n)+a(1) and a(n-1)+a(2) and a(n-2)+a(3)...

don't "just do it", apply some high faluting and uniformative mathematics, namely,

2. do it by induction, if you know what that is.

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HallsofIvy

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Start by looking at some examples: if a= 1, d= 3, then the sum is

1+ 4+ 7+ 10= (1+ 0(3))+ (1+ 1(3))+ (1+ 2(3))+ (1+ 3(3)) (here n= 4). I would think of that as (1+ 1+ 1+ 1)+ (0+ 1+ 2+ 3)(3). Obviously the first part of that is just 1 (a in general) added to itself n times: an while the other is 3 (d in general) times the sum of the series 1+ 2+ 3+...+ n-1. Do you know a formula for that. If not use Gauss' idea: Add 1+ 2+ 3+ ...+ (n-1) to (n-1)+...+ 3+ 2+ 1. If you add term by term you get

1+ (n-1)= n, 2+ (n-2)= n, ... up to (n-1)+ 1= n. In other words, every term is n and there are n-1 terms: the sum is n(n-1). Oops! That was adding the sum twice (once in the original order and then reversed) so we need to divide by 2. That sum is n(n-1)/2.

That is, we have a added to itself n times: an and we have d times n(n-1)/2:

an+ 2n(n-1)/2.

- #4

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Recall that t1 = a, and tn = a + (n - 1)d for an arithmetic sequence. Verify that the following formula represents the sum of n terms of an arithmetic series:

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

i don't know the summation notation yet and i am not given any values for "a" and "d" so i dont know how to actually "do" it

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matt grime

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but the values of a and d do not matter. the "just do it proof" is, relatively straight forward if you follow the hitns you've been given:

a+(a+d)+(a+2d)+...

and

a+(n-1)d+(a+(n-2)d)+...

add up the first and the last terms (which I;ll do for you: it is a and a+(n-1)d), the second and second to last terms (a+d and a+(n-2)d), the third and third to last terms and what do you get? (if the answers are different you'e done something wrong).

did you acutally think about the hints you were given? maths solutions don't suddenyl appear unless you are a genius, you need to think about them for a LONG time before it becomes clear.

a+(a+d)+(a+2d)+...

and

a+(n-1)d+(a+(n-2)d)+...

add up the first and the last terms (which I;ll do for you: it is a and a+(n-1)d), the second and second to last terms (a+d and a+(n-2)d), the third and third to last terms and what do you get? (if the answers are different you'e done something wrong).

did you acutally think about the hints you were given? maths solutions don't suddenyl appear unless you are a genius, you need to think about them for a LONG time before it becomes clear.

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