Animated Solution for Mathematics - Sequence and Series: If $S_1,S_2,S_3,\dots ,S_n$ are the sums of infinite geometric series whose first terms are $1,2,3,\dots ,n$ and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots ,\frac{1}{n+1}$ respectively, then find the values of $S_1^2+S_2^2+S_3^2+\cdots +S_{2n-1}^2$.

A Journey into Elegance

Welcome, future engineers and mathematicians. Today, we are going to dismantle a problem that, at first glance, might seem like a chaotic mess of infinite series.

You see a list of series $S_1,S_2,\dots ,S_n$, each with its own first term and common ratio, and your instinct might be to panic. But here is the secret: in JEE Advanced, complexity is often just a mask for a hidden, elegant pattern. Let us peel back that mask together.

The General Term as a Key

Instead of staring at the entire sequence, let us zoom in. We need to find a general expression for the $k$-th series, $S_k$.

Look at the pattern provided: the first terms are $1,2,3,\dots ,n$, so the first term of the $k$-th series is simply $a_k=k$. The common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots ,\frac{1}{n+1}$, which means the common ratio for our $k$-th series is $r_k=\frac{1}{k+1}$.

We have our building blocks. Now, we invoke the most powerful tool in our arsenal for infinite geometric progressions: the sum formula $S=\frac{a}{1-r}$.

But before we dive in, we must perform a sanity check. Does this series even converge? The condition for an infinite GP to converge is $|r|<1$.

Our ratio is $r_k=\frac{1}{k+1}$. Since $k\ge 1$, the denominator $k+1$ is at least $2$. Thus, $r_k\le \frac{1}{2}$, which is strictly less than $1$. The series converges beautifully. We are safe to proceed.

The Algebraic Unfolding

Now, let us substitute our values into the formula:

I know what you are thinking—this looks like a messy fraction. But let us take a breath and simplify the denominator.

$1-\frac{1}{k+1}$ is the same as $\frac{(k+1)-1}{k+1}$, which simplifies to $\frac{k}{k+1}$. Now, our expression for $S_k$ becomes:

When we divide by a fraction, we multiply by its reciprocal: $S_k=k\cdot \frac{k+1}{k}$. The $k$ in the numerator and the $k$ in the denominator cancel out with a satisfying 'pop'. We are left with $S_k=k+1$.

Isn't that incredible? All that complexity collapsed into a simple linear expression. This is the beauty of mathematics—the way it hides simplicity within layers of complexity.

The Summation Trap

The question asks for the sum $T=S_1^2+S_2^2+S_3^2+\cdots +S_{2n-1}^2$. Since we know $S_k=k+1$, we can rewrite this as:

Let us expand this to see what we are really summing: for $k=1$, we get $(1+1{)}^2=2^2$. For $k=2$, we get $(2+1{)}^2=3^2$. This continues until $k=2n-1$, where we get $(2n-1+1{)}^2=(2n{)}^2$.

So, $T=2^2+3^2+\cdots +(2n{)}^2$.

Here is where the trap lies. The standard formula for the sum of squares is:

Our series $T$ starts from $2^2$, but the formula starts from $1^2$. We must add $1^2$ to our series to use the formula, and then subtract it back out. So, $T=[1^2+2^2+3^2+\cdots +(2n{)}^2]-1^2$.

The Final Elegance

Now, we apply the formula with $N=2n$:

Simplifying the term in the bracket, $2(2n)+1=4n+1$. So, $T=\frac{2n(2n+1)(4n+1)}{6}-1$.

We can simplify the fraction by dividing the $2n$ and the $6$ by $2$, leaving us with $\frac{n(2n+1)(4n+1)}{3}-1$. To combine these, we use a common denominator of $3$:

And there you have it. We started with a daunting list of infinite series and ended with a clean, elegant algebraic expression. This is the essence of JEE preparation: not just memorizing formulas, but learning to see the structure beneath the surface.

Keep practicing, keep questioning, and most importantly, keep falling in love with the process.