Animated Solution for Mathematics - Binomial Theorem: If ${\sum }_{r=1}^9\left(\frac{r+3}{2^r}\right)\cdot {}^9{C}_r=\alpha {\left(\frac{3}{2}\right)}^9-\beta ,\alpha ,\beta \in \mathbb{N}$, then $(\alpha +\beta {)}^2$ is equal to

Taming the Binomial Beast

Imagine you are standing before a massive, intimidating structure—a complex summation. It looks like this: ${\sum }_{r=1}^9\left(\frac{r+3}{2^r}\right)\cdot {}^9{C}_r$.

At first glance, it feels like a chaotic mess of variables and powers. But in the world of JEE Advanced, we don't fear complexity; we dismantle it. The secret to this problem isn't brute force; it is the art of decomposition.

Divide and Conquer

When you see a numerator like $r+3$, your first instinct should be to separate the variables from the constants. We can rewrite the summation as two distinct parts:

By splitting them, we have turned one impossible problem into two manageable ones. This is the first step of a master strategist.

The Magic of the Identity

Now, look at Part 1. That $r$ in the numerator is the obstacle. It prevents us from using the standard binomial expansion directly.

But we have a powerful tool in our arsenal: the identity $r\cdot {}^n{C}_r=n\cdot {}^{n-1}{C}_{r-1}$. Think of this identity as a way to 'absorb' the $r$ into the combination, effectively reducing the power of the combination and shifting the index.

Applying this with $n=9$, we get:

Substituting this back into Part 1, we get ${\sum }_{r=1}^9\frac{9\cdot {}^8{C}_{r-1}}{2^r}$. We can pull out the constant $9$ and one factor of $\frac{1}{2}$ to align the index of the combination with the power of the denominator:

By letting $k=r-1$, this transforms into the beautiful expansion of $(1+\frac{1}{2}{)}^8$. Thus, Part 1 simplifies elegantly to:

Completing the Pattern

Now, let us turn our attention to Part 2. We have $3{\sum }_{r=1}^9{}^9{C}_r(\frac{1}{2}{)}^r$. This looks almost like $(1+\frac{1}{2}{)}^9$, but it is missing the $r=0$ term.

In mathematics, when a pattern is almost complete, we complete it ourselves. We add and subtract the $r=0$ term:

Since ${}^9{C}_0=1$ and $(\frac{1}{2}{)}^0=1$, the term we subtract is simply $1$. The summation now becomes $(1+\frac{1}{2}{)}^9$. So, Part 2 is:

The Final Assembly

We are at the finish line. Let us combine our two results:

To add these, we need a common base and exponent. We know that $\frac{9}{2}=3\cdot \frac{3}{2}$. Therefore, the first term becomes $3\cdot \frac{3}{2}\cdot (\frac{3}{2}{)}^8$, which is $3(\frac{3}{2}{)}^9$.

Now, the expression is simple:

Comparing this to $\alpha (\frac{3}{2}{)}^9-\beta$, we immediately see that $\alpha =6$ and $\beta =3$. The final calculation, $(\alpha +\beta {)}^2$, becomes: