The value of $1+\left(1+\frac{1}{3}\right) \frac{1}{4}+\left(1+\frac{1}{3}+\frac{1}{9}\right) \frac{1}{16}+\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right) \frac{1}{64}+\ldots \ldots$, is

We might think this expression seems overwhelming at first, but let's break it down to see the beautiful pattern hidden within.

$1+\left(1+\frac{1}{3}\right) \frac{1}{4}+\left(1+\frac{1}{3}+\frac{1}{9}\right) \frac{1}{16}+\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right) \frac{1}{64}+\ldots$

Key observations:

The denominators outside parentheses are: $4, 16, 64, \ldots$ which are $4^1, 4^2, 4^3, \ldots$

Inside each parenthesis, we have finite geometric series: $1, 1+\frac{1}{3}, 1+\frac{1}{3}+\frac{1}{9}, \ldots$

Each term inside uses powers of $\frac{1}{3}$: $\frac{1}{3^0}, \frac{1}{3^1}, \frac{1}{3^2}, \ldots$

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Instead of trying to sum this directly, let's group terms by their powers of $\frac{1}{3}$:

Terms with $\frac{1}{3^0} = 1$:

$1 \cdot 1 + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{16} + 1 \cdot \frac{1}{64} + \ldots$

Terms with $\frac{1}{3^1} = \frac{1}{3}$:

$\frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} \cdot \frac{1}{16} + \frac{1}{3} \cdot \frac{1}{64} + \ldots$

Terms with $\frac{1}{3^2} = \frac{1}{9}$:

$\frac{1}{9} \cdot \frac{1}{16} + \frac{1}{9} \cdot \frac{1}{64} + \frac{1}{9} \cdot \frac{1}{256} + \ldots$

This gives us:

$1\left(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots\right) + \frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots\right) + \frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+\ldots\right) + \ldots$

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Each parenthesis contains a geometric series with first term $a$ and common ratio $r = \frac{1}{4}$.

For an infinite geometric series: $\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$ (when $|r| < 1$)

First series: $S_1 = 1 + \frac{1}{4} + \frac{1}{16} + \ldots = \frac{1}{1-\frac{1}{4}} = \frac{4}{3}$

Second series: $S_2 = \frac{1}{3}\left(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots\right) = \frac{1}{3} \cdot \frac{\frac{1}{4}}{1-\frac{1}{4}} = \frac{1}{3} \cdot \frac{1/4}{3/4} = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$

Third series: $S_3 = \frac{1}{9}\left(\frac{1}{16} + \frac{1}{64} + \ldots\right) = \frac{1}{9} \cdot \frac{\frac{1}{16}}{1-\frac{1}{4}} = \frac{1}{9} \cdot \frac{1/16}{3/4} = \frac{1}{9} \cdot \frac{1}{12} = \frac{1}{108}$

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Our series has become:

$\frac{4}{3} + \frac{1}{9} + \frac{1}{108} + \ldots$

Let's check if this is geometric:

First term: $\frac{4}{3}$

Second term: $\frac{1}{9}$

Third term: $\frac{1}{108}$

Finding the common ratio:

$r = \frac{\frac{1}{9}}{\frac{4}{3}} = \frac{1}{9} \times \frac{3}{4} = \frac{1}{12}$

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We have a geometric series with:

First term: $a = \frac{4}{3}$

Common ratio: $r = \frac{1}{12}$

Sum of infinite geometric series:

$S = \frac{a}{1-r} = \frac{\frac{4}{3}}{1-\frac{1}{12}} = \frac{\frac{4}{3}}{\frac{11}{12}} = \frac{4}{3} \times \frac{12}{11} = \frac{48}{33} = \frac{16}{11}$

Therefore, the answer is $\frac{16}{11}$

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Key Takeaway: When faced with complex nested series, try rearranging terms by grouping similar components. This often reveals hidden geometric progressions that are much easier to sum!