Misanthropic

What was the equation for the sum of all terms in a arithmetic and geomatric sequence?
For example i saw question that stated find the sum of the first 50 terms in this sequence. I know it has a sigma bit couldn't, for the life of me, remember what it was.

hanngo2911

I just remember the formula of the sum of arithmatic sequence:
S= [(the first number + the final number)* the number of numbers]/2
Or the first number + (n-1)*distance
(n-1: the n-1 th term in the sequence)

DrSteve

The following little trick for an arithmetic series is better than memorizing a formula. See if you can explain what I'm doing:

1 + 2 + 3 + ... + 50
50 + 49 + 48 + ... + 1

51 + 51 + 51 + ... + 51 = 50(51) = 2550.

Finally divide by 2: 2550/2 = 1275.

If you want to know the formula, it is most easily remembered as (number of terms)*(average of first and last term) = 50(1 + 50)/2 = 1275.

I haven't yet seen problem on the SAT where you need the arithmetic series formula (but there are often "differences of large sums" problems)

Misanthropic

Thanks DrSteve
But do you know the rule for the geomatric series formula?

novelidea

Also, Dr Steve,

are there any other cool formulas we should know?

DrSteve

I have given a full list of "cool" formulas in my article "The Math Formulas You Should Memorize for the SAT." I've posted it on this forum several times - just do a search.

The geometric series formula is not one of these - there is no need to memorize this one.

That said, here it is:

Gn = g[(1-r^n)/(1-r)] where g is the first term of the series, r is the common ratio, and n is the number of terms.

The infinite geometric series has a simpler formula:

G = g/(1-r) if -1<r<1.

If r<-1 or r>1, then the infinite geometric series has no sum.

rspence

A simple proof for a geometric series:

Let S = 1 + r + r^2 + r^3 + ... (assuming S converges)

Then rS = r + r^2 + r^3 + ... = S - 1

Solving for S gives S = 1/(1-r) (where |r| < 1).

Afaloo

DrSteve, I tried to employ your approach to the following question_ which is by the way from the Test on January 2011_ but it didn't work. I'm sure there was a flaw in my calculation or something. So the question is: The first three terms of a sequence are given: 1/(1)(2), 1/(2)(3), 1/(3)(4)
The nth term of the sequence is
1/(n)(n+1), which is equal to
1/n - 1/n+1. What is the sum of the first 50 terms of this sequence?
Please explain it to me.

rspence

@Afaloo, the sequence is not arithmetic.

Clearly, 1/(n*(n+1)) = 1/n - 1/(n+1). The sum of the first 50 terms is

S = 1/(1*2) + 1/(2*3) + ... + 1/(50*51), which is equal to

S = (1/1 - 1/2) + (1/2 - 1/3) + ... + (1/50 - 1/51)

Here, notice that everything cancels out (-1/2 cancels out 1/2, etc.) leaving 1/1 - 1/51, or 50/51. This is known as a telescoping series.

Afaloo

@rspence, wow! Thank you so much, I really appreciate your help. I will post a couple of questions in a new thread, so I will be glad if you go through them once you have the time.

rspence

Sure, no problem.