Find the sum (assume |x| < 1/3):
1+3x+9x^2+27x^3+81x^4+...=?
Well, well, well! Looks like we have ourselves an infinite geometric series. Let's get ready to crunch some numbers, shall we?
The formula for the sum of an infinite geometric series is given by S = a / (1 - r), where "a" is the first term and "r" is the common ratio. In this case, our first term is 1, and our common ratio is 3x.
Now, we need to make sure that |r| < 1 in order for this series to converge. In other words, we want to make sure that |3x| < 1. But wait, what does that mean? It means that the absolute value of 3x must be less than 1. In this case, it also means that the absolute value of x must be less than 1/3.
Alright, back to business! Let's plug in our values into the formula and see what happens:
S = 1 / (1 - 3x)
But beware, this series might only converge within the specified range of |x| < 1/3. Outside of that, all bets are off!
So, my friend, there you have it! The sum of the series 1 + 3x + 9x^2 + 27x^3 + 81x^4 + ... (assuming |x| < 1/3) is 1 / (1 - 3x). Happy math-ing!
To find the sum of the infinite geometric series, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where:
- S is the sum of the series,
- a is the first term of the series,
- r is the common ratio.
In this case, the first term (a) is 1, and the common ratio (r) is 3x.
However, in order to use this formula, we need to ensure that the common ratio (3x) is between -1 and 1, as given by the condition |x| < 1/3.
Therefore, we need to find the range of x for which the series converges.
Since |x| < 1/3 and r = 3x, we have:
-1 < 3x < 1/3
Dividing all parts of the inequality by 3:
-1/3 < x < 1/9
So, for the series to converge, x must be between -1/3 and 1/9.
Now, we can use the formula to find the sum of the series:
S = 1 / (1 - 3x)
Therefore, the sum of the series 1+3x+9x^2+27x^3+81x^4+... is 1 / (1 - 3x), provided that -1/3 < x < 1/9.
To find the sum of this infinite geometric series, we can start by expressing it in the form of a geometric series:
1 + 3x + 9x^2 + 27x^3 + 81x^4 + ...
The common ratio (r) between consecutive terms can be found by dividing any term by the previous term:
r = (3x)/(1) = 3x
Now, we need to determine the condition for convergence, which is when the absolute value of the common ratio is less than 1:
|r| < 1
|3x| < 1
|3x| < 1/3
Given that |x| < 1/3, we can rewrite the inequality:
-1/3 < x < 1/3
Therefore, the series converges when |x| is less than 1/3.
To find the sum of this geometric series, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
Where:
S = sum of the series
a = first term of the series
r = common ratio
In this case, the first term (a) is 1 and the common ratio (r) is 3x.
Now, we can substitute these values into the formula and solve for the sum:
S = 1 / (1 - 3x)
Hence, the sum of the infinite geometric series 1 + 3x + 9x^2 + 27x^3 + 81x^4 + ... is 1 / (1 - 3x), when |x| < 1/3.
This is just a geometric series where the common ratio is 3x.
So, S = 1/(1-3x)