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What is the sum of the first n terms of a geometric sequence and where does it come from?

Recall that a geometric progression is a sequence (a_n) defined as follows

 

a_1 = a

a_n = a * r^(n - 1) for all integers n > 1

where a and r are some fixed numbers and r /= 0 is called the common ratio of a sequence.

 

Let's denote S_n as the sum of the first n terms of this sequence.

 

Case r = 1

Then all the terms are equal to a so

S_n = n*a

 

Case r /= 1

Then we have

S_n = a_1 + a_2 + … + a_n = a + a*r + … a*r^(n - 1) = a * (1 + r + … + r^(n – 1))

 

Now the formula for the difference between two n-th powers tells us that

r^n - 1 = r^n – 1^n = (r - 1) * (1 + r + … + r^(n-1))

and since r /= 1 we can divide both sides by r – 1 to have

1 + r + … + r ^(n - 1) = (r^n - 1) / (r – 1)

 

Finally, substituting this expression into the first equality we get

S_n = a * (r^n - 1) / (r – 1)

Kamil N. GCSE Maths tutor, A Level Maths tutor, 13 plus  Maths tutor,...

2 years ago

Answered by Kamil, an A Level Maths tutor with MyTutor


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