Animated Solution for Mathematics - Sequence and Series: If $A={\sum }_{n=1}^{\infty }\frac{1}{(3+(-1{)}^n{)}^n}$ and $B={\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{(3+(-1{)}^n{)}^n}$, then $\frac{A}{B}$ is equal to:

Welcome, future engineers. Today, we are going to peel back the layers of an infinite series problem that, at first glance, might look like a chaotic mess of powers and signs. But in the world of JEE Advanced, chaos is often just order in disguise.

Let us embark on this journey to find the ratio $\frac{A}{B}$.

Decoding the Heartbeat

The expression $A={\sum }_{n=1}^{\infty }\frac{1}{(3+(-1{)}^n{)}^n}$ is governed by a hidden rhythm. Look closely at the denominator base: $b=3+(-1{)}^n$.

This term is the heartbeat of our series. When $n$ is odd, $(-1{)}^n$ is $-1$, so the base becomes $3-1=2$. When $n$ is even, $(-1{)}^n$ is $1$, so the base becomes $3+1=4$.

This simple parity switch is the key to everything. It tells us that our series is not a single, uniform entity, but a composite of two distinct behaviors.

The Art of Decomposition

To solve this, we must separate the series into its odd and even components. Let us expand $A$ to see the pattern:

See how the terms alternate between powers of $2$ and powers of $4$? Let us group them.

The odd terms are $A_{\text{odd}}=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\dots$ and the even terms are $A_{\text{even}}=\frac{1}{4^2}+\frac{1}{4^4}+\frac{1}{4^6}+\dots$. Suddenly, the chaos vanishes. We are left with two beautiful, infinite Geometric Progressions (GPs).

The GP Engine

For $A_{\text{odd}}$, the first term $a=\frac{1}{2}$ and the common ratio $r=\frac{1}{4}$. Using the sum formula $S_{\infty }=\frac{a}{1-r}$, we get:

For $A_{\text{even}}$, the first term $a=\frac{1}{16}$ and the common ratio $r=\frac{1}{16}$. Applying the same formula:

Adding these together, the total value of $A$ is:

The Final Synthesis

Now, what about $B$? The series $B={\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{(3+(-1{)}^n{)}^n}$ is almost identical to $A$, but the numerator carries that alternating sign.

When $n$ is odd, the numerator is $-1$. When $n$ is even, it is $1$. This means $B=-A_{\text{odd}}+A_{\text{even}}$.

Substituting our previous results:

Finally, we calculate the ratio $\frac{A}{B}$:

The elegance of this result is in the cancellation. By keeping our fractions with a common denominator of $15$, the division becomes trivial.

Remember, in JEE Advanced, the most complex problems often yield to the simplest strategies. Always look for the parity, decompose the series, and let the Geometric Progression do the heavy lifting. You have mastered this! The final answer is $-\frac{11}{9}$.