<p>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.</p>
<p>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)</p>
<p>The following little trick for an arithmetic series is better than memorizing a formula. See if you can explain what I’m doing:</p>
<p>1 + 2 + 3 + … + 50
50 + 49 + 48 + … + 1</p>
<p>51 + 51 + 51 + … + 51 = 50(51) = 2550.</p>
<p>Finally divide by 2: 2550/2 = 1275.</p>
<p>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.</p>
<p>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)</p>
<p>Thanks DrSteve
But do you know the rule for the geomatric series formula?</p>
<p>Also, Dr Steve,</p>
<p>are there any other cool formulas we should know?</p>
<p>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. </p>
<p>The geometric series formula is not one of these - there is no need to memorize this one. </p>
<p>That said, here it is:</p>
<p>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.</p>
<p>The infinite geometric series has a simpler formula:</p>
<p>G = g/(1-r) if -1<r<1. </p>
<p>If r<-1 or r>1, then the infinite geometric series has no sum.</p>
<p>A simple proof for a geometric series:</p>
<p>Let S = 1 + r + r^2 + r^3 + … (assuming S converges)</p>
<p>Then rS = r + r^2 + r^3 + … = S - 1</p>
<p>Solving for S gives S = 1/(1-r) (where |r| < 1).</p>
<p>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.</p>
<p>@Afaloo, the sequence is not arithmetic.</p>
<p>Clearly, 1/(n*(n+1)) = 1/n - 1/(n+1). The sum of the first 50 terms is</p>
<p>S = 1/(1<em>2) + 1/(2</em>3) + … + 1/(50*51), which is equal to</p>
<p>S = (1/1 - 1/2) + (1/2 - 1/3) + … + (1/50 - 1/51)</p>
<p>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.</p>
<p>@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.</p>
<p>Sure, no problem.</p>