Find the sum of the infinite geometric series 16 + 4 + 1 + 1/4+ . . .
We can write the given infinite geometric series as
\[16 + 4 + 1 + \frac14 + \dotsb = \frac{16}{1} + \frac{16}{4} + \frac{16}{16} + \frac{16}{64} + \dotsb.\]To find the sum of this series, we want to find the value of $S = \frac{16}{1} + \frac{16}{4} + \frac{16}{16} + \frac{16}{64} + \dotsb$ and multiply it by $\frac{1}{1 - \frac14} = \frac43.$
Let
\[T = \frac{16}{4} + \frac{16}{16} + \frac{16}{64} + \dotsb.\]Then
\[T = \frac{16}{16} + \frac{16}{64} + \dotsb = \frac14 + \frac{1}{16} + \dotsb,\]so
\[T = \frac{1}{1 - \frac14} = \frac43.\]Then
\[S = \frac{16}{1} + T = \frac{16}{1} + \frac43 = \boxed{\frac{64}{3}}.\]
To find the sum of an infinite geometric series, you can use the formula:
S = a / (1 - r)
Where:
S = sum of the series
a = first term of the series
r = common ratio
In your case, the first term is 16, and the common ratio is 1/4. Plugging these values into the formula:
S = 16 / (1 - 1/4)
Simplifying the denominator:
S = 16 / (3/4)
To divide by a fraction, we multiply by its reciprocal:
S = 16 * (4/3)
Multiplying the numerator:
S = 64 / 3
So, the sum of the infinite geometric series 16 + 4 + 1 + 1/4 + ... is 64/3.
To find the sum of an infinite geometric series, you need to determine whether the series converges or diverges.
In this case, we have a geometric series with a common ratio (r) of 1/4. To determine if the series converges, we need to analyze the magnitude of the common ratio |r|.
Since |1/4| is less than 1, the series will converge. Now, let's 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 our case, the first term (a) is 16, and the common ratio (r) is 1/4.
S = 16 / (1 - 1/4)
To simplify, we can find a common denominator:
S = 16 / (4/4 - 1/4)
S = 16 / (3/4)
S = 16 * 4/3
S = 64/3
So, the sum of the given infinite geometric series 16 + 4 + 1 + 1/4 + ... is 64/3.