The Formula for 1 + a + a^2 + a^3 + … + a^n
The formula for 1 + a + a^2 + a^3 + … + a^n is a well-known mathematical expression that has numerous applications in various fields, including algebra, geometry, and calculus. In this article, we will explore the formula, its derivation, and some examples to illustrate its usage.
The Formula
The formula for 1 + a + a^2 + a^3 + … + a^n is given by:
1 + a + a^2 + a^3 + … + a^n = (a^(n+1) - 1) / (a - 1)
where 'a' is a real number and 'n' is a positive integer.
Derivation
To derive the formula, let's consider the geometric series:
1 + a + a^2 + a^3 + … + a^n
We can multiply both sides of the equation by 'a' to get:
a + a^2 + a^3 + … + a^(n+1)
Now, subtract the original series from the new series:
a^(n+1) - 1 = (a + a^2 + a^3 + … + a^n) - (1 + a + a^2 + a^3 + … + a^n)
Simplifying the equation, we get:
a^(n+1) - 1 = a(1 + a + a^2 + a^3 + … + a^(n-1)) - (1 + a + a^2 + a^3 + … + a^n)
Factoring out (1 + a + a^2 + a^3 + … + a^(n-1)) from both terms, we get:
a^(n+1) - 1 = (a - 1)(1 + a + a^2 + a^3 + … + a^(n-1))
Dividing both sides by (a - 1), we arrive at the formula:
1 + a + a^2 + a^3 + … + a^n = (a^(n+1) - 1) / (a - 1)
Examples
Let's illustrate the formula with some examples:
Example 1 Find the sum of 1 + 2 + 2^2 + 2^3 + … + 2^5.
Using the formula, we get:
1 + 2 + 2^2 + 2^3 + … + 2^5 = (2^6 - 1) / (2 - 1) = 63
Example 2 Find the sum of 1 + x + x^2 + x^3 + … + x^4, where x = 3.
Using the formula, we get:
1 + 3 + 3^2 + 3^3 + … + 3^4 = (3^5 - 1) / (3 - 1) = 40
Conclusion
In conclusion, the formula for 1 + a + a^2 + a^3 + … + a^n has numerous applications in mathematics and other fields. The formula is derived by manipulating the geometric series and factoring out the common terms. We have illustrated the formula with some examples to demonstrate its usage.
