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Malhar Rajpal Most high school students will have heard of the infinite geometric series formula as well as the sum of an arithmetic sequence formula. Most of us just accept these blindly without an explanation as to why they truly work. However, today we will prove both the formulas infinite and finite geometric series sums as well as the formula for the sum of an arithmetic sequence! For those who haven’t encountered these terms before, an arithmetic sequence is a sequence of numbers such that the difference between consecutive terms is constant. The nth sum of an arithmetic progression is thus the sum of all the numbers in the sequence up to the nth term of the sequence. Evidently, since the difference between consecutive terms is constant, the numbers will eventually be either very large and positive or very low and negative and the infinite sum of an arithmetic sequence is thus divergent. An infinite geometric series is the sum of an infinite number of terms in a sequence wherein there is a constant ratio between successive terms (sum of infinite values in a geometric sequence). It is important to note than an infinite geometric series only converges if the absolute value of the common ratio r is less than 1. As the name suggests, a finite geometric series is the sum of a finite number of terms (up to the denoted nth term) of a geometric sequence. Arithmetic Series formula proof: The formula for an arithmetic series, adding up each of the first n terms of an arithmetic progression is given by: where S_n the sum of the first n terms of an arithmetic sequence, a is the first term of the sequence, n is the number of terms we want in the sum, and d is the common difference between successive terms in the arithmetic sequence. To prove this formula, we must first look at the base definition of an arithmetic series which is adding up the first n terms of an arithmetic sequence: S_n=a+(a+d)+(a+2d)+(a+3d)+...+(a+(n-2)d)+(a+(n-1)d), or Sn=(a+(n-1)d)+(a+(n-2)d)+...+(a+3d)+(a+2d)+(a+d)+a (which is the same as above but backwards. If we add the first term of the first S_n formula with the first term of the second S_n formula, and the second term of the first S_n formula with the second term of the second Sn formula, and so on, we see that each term adds up to 2a+(n-1)d: For example: a+(a+(n-1)d) = 2a+(n-1)d and (a+d)+(a+(n-2)d) = 2a+(n-1)d… Hence by adding both the S_n together, we get: 2S_n=(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d) Since we are adding the first n terms of the arithmetic sequence, there are thus n occurrences of 2a+(n-1)d in the formula for 2S_n. Thus: 2S_n=n(2a+(n-1)d) And by dividing two from both sides, we reach our desired formula: S_n=(n/2)(2a+(n-1)d) And the arithmetic series formula is proven. Finite Geometric Series Proof: The formula for the sum of a finite geometric series is: where S_n in this case denotes the sum of the first n terms of a geometric sequence. r is the common ratio between consecutive terms and n is the number of terms we want to sum. To prove this formula, let’s look at the defining formula for a finite geometric sequence, which includes summing up the n terms of the sequence: Sn=a+ar+ar^2+...+ar^(n-2)+ar^(n-1) We can create another formula by multiplying each term by r: rSn=ar+ar^2+....+ar^(n-2)+ar^(n-1)+ar^n The difference between these two formulas is that the formula for S_n has the term a while the formula for rS_n has the term ar^n. Both the formulas have all the other terms, ar+ar^2+...+ar^(n-1). Because of this, we can subtract rS_n from S_n to get: S_n-rS_n=a-ar^n. Factoring out both sides: S_n(1-r)=a(1-r^n) Dividing both sides by 1-r (which is possible since we don’t consider r=1 in this case) to get our desired formula: S_n=a(1-rn)/(1-r) , r≠1 And we have proven the formula for a finite geometric series! Infinite Geometric Series Proof: The textbook formula for the sum of an infinite geometric series is: S_∞=a/(1-r), -1<r<1 To determine the proof, we can evaluate the S_n formula for a finite geometric series as n tends to ∞. We thus get: The limit can only be solved if -1<r<1 since if r doesn’t fall within this range, the r^n term diverges as n tends to ∞. However, if r falls within the range -1<r<1, as n approaches ∞, the r^n term approaches 0. Thus we get our desired formula: And the final formula for this article is proven!
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