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:

S=140+341+642+1043+1544+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:

Tn=n(n+1)4nT_n = \frac{n(n+1)}{4^n}

Where TnT_n is the nth term in the series.

We are asked to find S3\sqrt[3]{S}, which is the cube root of the sum of the series SS.

Step 1: Sum of the Series

To find the sum of the series SS, we recognize the general term as:

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

This can be split into two simpler geometric series:

S=n=0n24n+n=0n4nS = \sum_{n=0}^{\infty} \frac{n^2}{4^n} + \sum_{n=0}^{\infty} \frac{n}{4^n}

Series 1: n=0n4n\sum_{n=0}^{\infty} \frac{n}{4^n}

The sum of this series is known to be:

n=0nxn=x(1x)2\sum_{n=0}^{\infty} \frac{n}{x^n} = \frac{x}{(1-x)^2} For x=14x = \frac{1}{4}, this becomes:

n=0n4n=14(114)2=14(34)2=14916=49\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: n=0n24n\sum_{n=0}^{\infty} \frac{n^2}{4^n}

The sum of this series is known to be:

n=0n2xn=x(1+x)(1x)3\sum_{n=0}^{\infty} \frac{n^2}{x^n} = \frac{x(1+x)}{(1-x)^3} For x=14x = \frac{1}{4}, this becomes:

n=0n24n=14(1+14)(114)3=14×54(34)3=5162764=5×6416×27=320432=4054=2027\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=49+2027=1227+2027=3227S = \frac{4}{9} + \frac{20}{27} = \frac{12}{27} + \frac{20}{27} = \frac{32}{27}

Step 3: Cube Root

Finally, we find S3\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, S3=43\sqrt[3]{S} = \frac{4}{3}.

Final Answer:

43\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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