Animated Solution for Mathematics - Sequence and Series: The value of $\frac{1\times 2^2+2\times 3^2+\cdots +100\times (101{)}^2}{1^2\times 2+2^2\times 3+\cdots +100^2\times 101}$ is

Unlocking the Series

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of arithmetic. You see a massive fraction, a long string of products in the numerator, and an equally intimidating series in the denominator.

It is designed to make you panic. But here is the secret: in the world of JEE, intimidation is just a mask for a beautiful, hidden structure. Let us peel back that mask together.

Decoding the DNA

First, let us give our monsters names. Let the numerator be $N$ and the denominator be $D$. Our goal is to find the ratio $\frac{N}{D}$.

Instead of staring at the whole thing, let us zoom in on the $r$-th term. Look at the numerator: $1\times 2^2,2\times 3^2,\dots$. The pattern is clear! The first number is $r$, and the second is $(r+1{)}^2$.

So, the general term is $T_r=r(r+1{)}^2$. Now, look at the denominator: $1^2\times 2,2^2\times 3,\dots$. Here, the square has shifted! The general term is $T_r^{\prime }=r^2(r+1)$. We have successfully decoded the DNA of this series.

The Power of Sigma

Now that we have our general terms, we need to sum them up. But first, we must expand them into simple polynomials.

For the numerator, $T_r=r(r^2+2r+1)=r^3+2r^2+r$. For the denominator, $T_r^{\prime }=r^2(r+1)=r^3+r^2$.

Now, we can use the power of sigma notation. The numerator $N$ becomes ${\sum }_{r=1}^{100}(r^3+2r^2+r)$, which splits into $\sum r^3+2\sum r^2+\sum r$. The denominator $D$ becomes ${\sum }_{r=1}^{100}(r^3+r^2)$, which is $\sum r^3+\sum r^2$.

This is where the magic happens. We are no longer dealing with a list of numbers; we are dealing with the fundamental building blocks of algebra.

The Art of Simplification

We know our standard formulas:

Many students would rush to calculate these values for $n=100$ right now. Do not do that! That is the trap.

Instead, let us factor out the common term $\frac{n(n+1)}{2}$ from both $N$ and $D$. When we do this, the ratio $\frac{N}{D}$ simplifies beautifully. The common factor cancels out entirely, leaving us with a much more manageable expression:

The Final Calculation

Now, and only now, do we substitute $n=100$. We calculate the pieces:

The numerator becomes $5050+2(67)+1=5185$. The denominator becomes $5050+67=5117$.

We are left with $\frac{5185}{5117}$. A quick check for common factors reveals that both are divisible by $17$. Dividing both by $17$, we get the final result:

And there it is! The monster is defeated, the structure is revealed, and the answer is clear. Remember, in JEE, it is not about brute force; it is about finding the elegant path through the complexity.