Animated Solution for Mathematics - Binomial Theorem: If ${\sum }_{r=1}^{30}\frac{r^2{(}^{30}{C}_r{)}^2}{{}^{30}{C}_{r-1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of factorials and powers.

We are given the expression:

It looks intimidating, doesn't it? But remember, in the world of mathematics, complexity is often just a mask for underlying simplicity. Let us peel back that mask together.

The Surgical Strike

Our first step is to simplify the ratio of consecutive binomial coefficients. We see the term $\frac{{}^{30}{C}_r}{{}^{30}{C}_{r-1}}$ embedded within our summation.

The identity $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$ is our best friend here. For $n=30$, this becomes $\frac{31-r}{r}$.

By substituting this into our original expression, we can cancel out the $r$ in the denominator with one of the $r$ terms in the numerator. The expression transforms into:

The chaos is already beginning to subside.

The Magic of Absorption

Now we face the term $r{\cdot }^{30}{C}_r$. This is where the absorption identity, $r{\cdot }^n{C}_r=n{\cdot }^{n-1}{C}_{r-1}$, shines.

It allows us to absorb the $r$ into the binomial coefficient, reducing the upper index from $30$ to $29$. Applying this, our sum becomes:

We have successfully reduced the complexity of the binomial coefficient itself.

The Perspective Shift

To make this sum truly transparent, we perform an index shift. Let $k=r-1$. As $r$ goes from $1$ to $30$, $k$ goes from $0$ to $29$.

The term $(31-r)$ becomes $(31-(k+1))$, which is $(30-k)$. Our expression is now:

This is the moment of clarity. We can split this into two standard summations:

The Grand Finale

We know that \sum_{k=0}^{n} ^{n}C_{k} = 2^{n} and ${\sum }_{k=0}^{n}k{\cdot }^n{C}_k=n\cdot 2^{n-1}$. For $n=29$, these are $2^{29}$ and $29\cdot 2^{28}$, respectively.

Substituting these values, we get:

By factoring out $2^{28}$, we get $30\cdot 2^{28}[60-29]$, which simplifies to $30\cdot 31\cdot 2^{28}$.

Adjusting the power of $2$ to match $2^{29}$, we get:

By direct comparison, $\alpha =465$. We have conquered the beast.

Remember, the path to the answer is not through brute force, but through the elegant application of the identities you have mastered. Keep practicing, and you will find that even the most terrifying problems have a beautiful, logical heart.