Prove that; Sn = a(r^n - 1)/1
To prove the equation Sn = a(r^n - 1)/1 for a geometric series, we can use the formula for the sum of a geometric series:
Sn = a(1 - r^n) / (1 - r) [Formula for the sum of a geometric series]
Let's simplify this formula step by step:
1. Multiply both sides by (1 - r) to get rid of the fraction:
Sn(1 - r) = a(1 - r^n)
2. Expand the left side using the distributive property:
Sn - Snr = a - ar^n
3. Group the terms containing Sn on the left side:
Sn - a = ar^n - Snr
4. Add a to both sides:
Sn = ar^n - Snr + a
5. Rearrange the terms:
Sn + Snr = ar^n + a
6. Factor out Sn on the left side:
Sn(1 + r) = ar^n + a
7. Divide both sides by (1 + r):
Sn = (ar^n + a) / (1 + r)
8. Simplify the fraction:
Sn = a(r^n + 1) / (1 + r)
Therefore, the correct formula for Sn is Sn = a(r^n + 1) / (1 + r), not Sn = a(r^n - 1) / 1.
To prove the formula Sn = a(r^n - 1) / (r - 1), we need to use the formulas for the sum of a geometric series and the formula for the sum of a finite geometric series.
The formula for the sum of a geometric series is given by:
Sn = a * (1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms.
To derive this formula, we can start by considering the partial sum of a geometric series:
Sn = a + ar + ar^2 + ... + ar^(n-1)
Multiply both sides of the equation by r:
rSn = ar + ar^2 + ... + ar^(n-1) + ar^n
Subtract the original equation from the one multiplied by r:
rSn - Sn = ar^n - a
Factor out Sn:
Sn(r - 1) = a(r^n - 1)
Divide both sides by (r - 1):
Sn = a(r^n - 1) / (r - 1)
Therefore, we have proved the formula Sn = a(r^n - 1) / (r - 1).