Derivations of series equations

[fred2028]

So we've been taught the following four equations for arithmetic and geometric series. However, I asked the teacher and he wouldn't explain how they were gotten ... I find things easier to learn if I understand them instead of just memorizing. The equations are:

Infinite arithmetic series: Sn = n/2[2a + (n-1)d]
Finite arithmetic series: Sn = n/2(a + Tn)
Geometric series: Sn = [2(r^n - 1) / (r - 1)]

OR

Sn = [a(1-r^n) / (1-r)]

Where a is the first term (t1), Tn is the last term, and n is the term number. D is the common difference.
So I was wondering: How are these 4 equations derived?

 

[tkhunny]

Infinite arithmetic series: Sn = n/2[2a + (n-1)d]
Finite arithmetic series: Sn = n/2(a + Tn)
Geometric series: Sn = [2(r^n - 1) / (r - 1)]
OR
Sn = [a(1-r^n) / (1-r)]

I havn't had a laugh like that in a while. Are you SURE you were given a formula for an INFINITE arithmetic series? I'm thinking they are both FINITE.

My Rules of Learning:
1) Memorize Meaningless Formulas - Maybe Good Enough, but not likely
2) Have Someone Explain them a bit - Possibly Good Enough
3) Derive Them Yourself - Almost always good enough.
4) Play With them until they are friends of yours - Great!

These are a little tedious, but not too difficult. I'll get you started.

Finite Arithmetic

First Term = a
Second Term = a + d
Third Term = a + 2d
...
nth Term = a + (n-1)d

Add them all up

a + (a+d) + (a + 2d) + ... + (a+(n-1)d) =
n*a + d + 2d + 3d + ... + (n-1)d =
n*a + d[1 + 2 + 3 + ... + (n-1)] = ??

How do you add up all those values inside the square brackets?

I will leave it as an exercise.

Finite Geometric

First Term = a
Second Term = ar
Third Term = ar^2
...
nth Term = ar^(n-1)


a + ar + ar^2 + ... + ar^(n-1) =
a[1 + r + r^2 + ... + r^(n-1)] = ??

How do you add up all that stuff inside the square brackets?

Hint:

If 1 + r + r^2 + ... + r^(n-1) = S
and r(1 + r + r^2 + ... + r^(n-1)) = (r + r^2 + r^3 + ... + r^n)= Sr
What can we conclude about S?

I'll leave it as an exercise.

 

[soroban]

Hello, Fred!

Your formulas are incorrect . . .


Arithmetic sequence: .first term \(\displaystyle a\), common difference \(\displaystyle d\)

. . The \(\displaystyle n^{th}\) term is: \(\displaystyle \L\:a_n\:=\:a\,+\,(n\,-\,1)d\)

. . The sum of the first \(\displaystyle n\) tems is: \(\displaystyle \L\:S_n\;=\;\frac{n}{2}\left[2a\,+\,(n-1)d\right]\)


Geometric sequence: . first term \(\displaystyle a\), common ratio \(\displaystyle r.\)

. . The \(\displaystyle n^{th}\) term is: \(\displaystyle \L\:a_n \:=\:ar^{n-1}\)

. . The sum of the first \(\displaystyle n\) terms is: \(\displaystyle \L\:S_n \;=\;a\,\frac{1\,-\,r^n}{1\,-\,r}\)

 

[TchrWill]

So we've been taught the following four equations for arithmetic and geometric series. However, I asked the teacher and he wouldn't explain how they were gotten ... I find things easier to learn if I understand them instead of just memorizing. The equations are:

Infinite arithmetic series: Sn = n/2[2a + (n-1)d]
Finite arithmetic series: Sn = n/2(a + Tn)
Geometric series: Sn = [2(r^n - 1) / (r - 1)]

OR

Sn = [a(1-r^n) / (1-r)]

Where a is the first term (t1), Tn is the last term, and n is the term number. D is the common difference.
So I was wondering: How are these 4 equations derived?

The arithmetic series is the easiest to derive.

n......1......2......3......4......5......6...........term number n
N.....3......9......15....21....27.....33.........nth term Nn
S.....3.....12.....27....48....75....108........sum of consecutive terms
Diff.....9.....12.....15.....21....33
Diff........6.......6.......6.....6

The sum sequence is a finite difference sequence with the 2nd differences constant at 6. WIth 2nd differences constant, the general expression for the sum of the terms of the arithmetic sequence is of the form S = ax^2 + bx + c

Using the data, we can write

a(1)^2 + b(1) + c = 3 or a + b + c = 3
a(2)^2 + b(2) + c = 12 or 4a + 2b + c = 12
a(3)^2 + b(3) + c = 27or 9a + 3b + c = 27

Solving, a = 3, b = 0 and c = 0 making S = 3n^2