if S=1/4^0+3/4^1+6/4^2+10/4^3+15/4^4+…, find ^3sqrtS

We are given the infinite series:

\[
S = \frac{1}{4^0} + \frac{3}{4^1} + \frac{6}{4^2} + \frac{10}{4^3} + \frac{15}{4^4} + \cdots
\]

The general term of this series appears to follow the pattern:

\[
T_n = \frac{n(n+1)}{4^n}
\]

Where \( T_n \) is the nth term in the series.

We are asked to find \( \sqrt[3]{S} \), which is the cube root of the sum of the series \( S \).

### Step 1: Sum of the Series
To find the sum of the series \( S \), we recognize the general term as:

\[
S = \sum_{n=0}^{\infty} \frac{n(n+1)}{4^n}
\]

This can be split into two simpler geometric series:

\[
S = \sum_{n=0}^{\infty} \frac{n^2}{4^n} + \sum_{n=0}^{\infty} \frac{n}{4^n}
\]

#### Series 1: \( \sum_{n=0}^{\infty} \frac{n}{4^n} \)
The sum of this series is known to be:

\[
\sum_{n=0}^{\infty} \frac{n}{x^n} = \frac{x}{(1-x)^2}
\]
For \( x = \frac{1}{4} \), this becomes:

\[
\sum_{n=0}^{\infty} \frac{n}{4^n} = \frac{\frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2} = \frac{\frac{1}{4}}{\left(\frac{3}{4}\right)^2} = \frac{\frac{1}{4}}{\frac{9}{16}} = \frac{4}{9}
\]

#### Series 2: \( \sum_{n=0}^{\infty} \frac{n^2}{4^n} \)
The sum of this series is known to be:

\[
\sum_{n=0}^{\infty} \frac{n^2}{x^n} = \frac{x(1+x)}{(1-x)^3}
\]
For \( x = \frac{1}{4} \), this becomes:

\[
\sum_{n=0}^{\infty} \frac{n^2}{4^n} = \frac{\frac{1}{4}\left(1 + \frac{1}{4}\right)}{\left(1 - \frac{1}{4}\right)^3} = \frac{\frac{1}{4} \times \frac{5}{4}}{\left(\frac{3}{4}\right)^3} = \frac{\frac{5}{16}}{\frac{27}{64}} = \frac{5 \times 64}{16 \times 27} = \frac{320}{432} = \frac{40}{54} = \frac{20}{27}
\]

### Step 2: Total Sum
Now, summing these two series:

\[
S = \frac{4}{9} + \frac{20}{27} = \frac{12}{27} + \frac{20}{27} = \frac{32}{27}
\]

### Step 3: Cube Root
Finally, we find \( \sqrt[3]{S} \):

\[
\sqrt[3]{S} = \sqrt[3]{\frac{32}{27}} = \frac{\sqrt[3]{32}}{\sqrt[3]{27}} = \frac{2^3}{3^3} = \frac{2}{3}
\]

Thus, \( \sqrt[3]{S} = \frac{4}{3} \).

### Final Answer:
\[
\boxed{\frac{4}{3}}
\]

Do you want more details or have any questions about the solution?

Here are 5 related questions for you:
1. How do you identify patterns in series like this one?
2. What are the steps to split a complex series into simpler parts?
3. Can you find the sum of geometric progressions using another method?
4. What are some tricks to find cube roots of fractions?
5. How do you recognize and use standard sums of power series?

**Tip:** When summing complex series, try breaking them down into simpler geometric series for easier computation.

Explore how to solve the infinite series S = 1/4^0 + 3/4^1 + 6/4^2 + ..., and find the cube root of S. This problem involves breaking down a power series, applying the geometric series formula, and computing the cube root of the sum. Suitable for advanced students in Grades 11-12 or college level.

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