Animated Solution for Mathematics - Sequence and Series: If $0<x<1$ and $y=\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\dots$, then the value of $e^{1+y}$ at $x=\frac{1}{2}$ is:

A Journey Through Patterns

Imagine you are standing on the edge of an infinite sum, looking at a sequence of numbers that seems to grow in complexity. You see:

At first glance, it looks like a chaotic mess of fractions and powers. But in the world of JEE Advanced, chaos is just a pattern waiting to be decoded. Let us embark on a journey to simplify this expression and find the value of $e^{1+y}$ at $x=\frac{1}{2}$.

Decoding the General Term

To conquer any infinite series, we must find its heartbeat—the general term $T_n$. Look at the coefficients: $\frac{1}{2},\frac{2}{3},\frac{3}{4}$.

For a term with $x^n$, the coefficient is clearly $\frac{n-1}{n}$. Thus, our general term is $T_n=\frac{n-1}{n}{x}^n$.

Since the series begins with $x^2$, our index $n$ starts at $2$ and marches toward infinity. We can write this as:

This compact form is our map.

The Art of Splitting

Now, we face a hurdle. We cannot evaluate this sum directly. But here is the secret: algebra is the art of rewriting things in more useful ways.

Let us rewrite the fraction $\frac{n-1}{n}$ as $1-\frac{1}{n}$. Substituting this back into our summation, we get:

By distributing $x^n$, we can split this into two separate, manageable sums:

The GP and the Logarithm

Let us tackle the first part, $S_1={\sum }_{n=2}^{\infty }{x}^n=x^2+x^3+x^4+\dots$. This is a classic infinite Geometric Progression!

The first term $a$ is $x^2$, and the common ratio $r$ is $x$. Since $0<x<1$, the sum is simply:

Now for the second part, $S_2={\sum }_{n=2}^{\infty }\frac{x^n}{n}=\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots$. This reminds us of the Maclaurin series for $\ln (1-x)$, which is $\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$.

If we multiply by $-1$, we get $-\ln (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots$. Our $S_2$ is missing the first term, $x$. So:

The Final Assembly

Combining these, we have $y=\frac{x^2}{1-x}-(-\ln (1-x)-x)$. Simplifying this, we get:

By taking a common denominator for the algebraic terms:

Now, substitute $x=\frac{1}{2}$:

Finally, our target is $e^{1+y}$. Substituting $y$, we get:

We have arrived at the answer: $\frac{1}{2}{e}^2$. Remember, the complexity of a problem is often just a mask for a beautiful, simple structure. Keep practicing, and you will see these patterns everywhere!