Math Problem Statement
if S=1/4^0+3/4^1+6/4^2+10/4^3+15/4^4+…, find ^3sqrtS
Solution
We are given the infinite series:
The general term of this series appears to follow the pattern:
Where is the nth term in the series.
We are asked to find , which is the cube root of the sum of the series .
Step 1: Sum of the Series
To find the sum of the series , we recognize the general term as:
This can be split into two simpler geometric series:
Series 1:
The sum of this series is known to be:
For , this becomes:
Series 2:
The sum of this series is known to be:
For , this becomes:
Step 2: Total Sum
Now, summing these two series:
Step 3: Cube Root
Finally, we find :
[ \sqrt[3]{S} = \sqrt[3]{\frac{32}{27}} = \frac{\sqrt[3]{32}}{\sqrt[3]{27}} = \frac{2^3}{3^3} = \frac{2}{3} ]
Thus, .
Final Answer:
Do you want more details or have any questions about the solution?
Here are 5 related questions for you:
- How do you identify patterns in series like this one?
- What are the steps to split a complex series into simpler parts?
- Can you find the sum of geometric progressions using another method?
- What are some tricks to find cube roots of fractions?
- 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.
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Math Problem Analysis
Mathematical Concepts
Series
Geometric Series
Power Series
Cube Root
Formulas
T_n = \frac{n(n+1)}{4^n}
\sum_{n=0}^{\infty} \frac{n}{x^n} = \frac{x}{(1-x)^2}
\sum_{n=0}^{\infty} \frac{n^2}{x^n} = \frac{x(1+x)}{(1-x)^3}
S = \sum_{n=0}^{\infty} \frac{n(n+1)}{4^n}
\sqrt[3]{S}
Theorems
Geometric Series Formula
Sum of Power Series
Suitable Grade Level
Grade 11-12 / College level
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