1. Introduction: Key Concepts and Formulas

This problem requires a strong understanding of the Binomial Theorem and its properties. Specifically, we will use:

• Binomial Theorem: The expansion of $(x+y)^n$ is given by $(x+y)^n = \sum_{r=0}^n {}^n C_r x^{n-r} y^r$ A common special case is $(1+x)^n = \sum_{r=0}^n {}^n C_r x^r$.
• Property of Binomial Coefficients: A crucial identity for simplifying terms involving $r \cdot {}^n C_r$ is $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$ This identity is derived from the definition ${}^n C_r = \frac{n!}{r!(n-r)!}$. So, $r \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r)!} = n \cdot \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!} = n \cdot {}^{n-1} C_{r-1}$.

2. Problem Setup and Splitting the Summation

We are asked to evaluate the sum: $S = \sum_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r$ The first step in simplifying such summations is often to separate terms in the numerator. In this case, we have $(r+3)$ in the numerator, so we can split the sum into two parts: $S = \sum_{r=1}^9 \frac{r}{2^r} \cdot{ }^9 C_r + \sum_{r=1}^9 \frac{3}{2^r} \cdot{ }^9 C_r$ Let's call the first sum $S_1$ and the second sum $S_2$. $S_1 = \sum_{r=1}^9 \frac{r}{2^r} \cdot{ }^9 C_r$ $S_2 = \sum_{r=1}^9 \frac{3}{2^r} \cdot{ }^9 C_r$

3. Evaluating the First Part of the Summation ($S_1$)

For $S_1$, we have a term $r \cdot {}^9 C_r$. This immediately suggests using the property $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$. Here, $n=9$. $S_1 = \sum_{r=1}^9 \frac{1}{2^r} \left(9 \cdot {}^{9-1} C_{r-1}\right)$ $S_1 = 9 \sum_{r=1}^9 \frac{1}{2^r} \cdot {}^{8} C_{r-1}$ To fit the form of a binomial expansion $(1+x)^n = \sum_{k=0}^n {}^n C_k x^k$, we need the exponent of $\frac{1}{2}$ to match the lower index of the binomial coefficient. Let $k = r-1$. When $r=1$, $k=0$. When $r=9$, $k=8$. Also, we can write $\frac{1}{2^r} = \frac{1}{2 \cdot 2^{r-1}} = \frac{1}{2} \left(\frac{1}{2}\right)^{r-1}$. $S_1 = 9 \sum_{k=0}^8 \frac{1}{2} \left(\frac{1}{2}\right)^k \cdot {}^{8} C_k$ $S_1 = \frac{9}{2} \sum_{k=0}^8 {}^{8} C_k \left(\frac{1}{2}\right)^k$ Now, the summation is precisely the binomial expansion of $\left(1+\frac{1}{2}\right)^8$. $S_1 = \frac{9}{2} \left(1+\frac{1}{2}\right)^8 = \frac{9}{2} \left(\frac{3}{2}\right)^8$

4. Evaluating the Second Part of the Summation ($S_2$)

For $S_2$, we have a constant '3' in the numerator, which can be factored out. $S_2 = \sum_{r=1}^9 \frac{3}{2^r} \cdot{ }^9 C_r$ $S_2 = 3 \sum_{r=1}^9 {}^{9} C_r \left(\frac{1}{2}\right)^r$ This sum looks very similar to the binomial expansion of $\left(1+\frac{1}{2}\right)^9$. The full expansion is $\left(1+\frac{1}{2}\right)^9 = \sum_{r=0}^9 {}^{9} C_r \left(\frac{1}{2}\right)^r$. Notice that our sum $S_2$ starts from $r=1$ instead of $r=0$. This means the $r=0$ term is missing from $S_2$. The $r=0$ term for the expansion of $\left(1+\frac{1}{2}\right)^9$ is ${}^{9} C_0 \left(\frac{1}{2}\right)^0 = 1 \cdot 1 = 1$. Therefore, we can write: $\sum_{r=1}^9 {}^{9} C_r \left(\frac{1}{2}\right)^r = \left(\sum_{r=0}^9 {}^{9} C_r \left(\frac{1}{2}\right)^r\right) - {}^{9} C_0 \left(\frac{1}{2}\right)^0$ $\sum_{r=1}^9 {}^{9} C_r \left(\frac{1}{2}\right)^r = \left(1+\frac{1}{2}\right)^9 - 1 = \left(\frac{3}{2}\right)^9 - 1$ Substituting this back into the expression for $S_2$: $S_2 = 3 \left[ \left(\frac{3}{2}\right)^9 - 1 \right]$

5. Combining the Results and Simplification

Now we combine $S_1$ and $S_2$ to find the total sum $S$: $S = S_1 + S_2$ $S = \frac{9}{2} \left(\frac{3}{2}\right)^8 + 3 \left[ \left(\frac{3}{2}\right)^9 - 1 \right]$ Let's simplify the expression: $S = \frac{3^2}{2} \cdot \frac{3^8}{2^8} + 3 \cdot \frac{3^9}{2^9} - 3$ $S = \frac{3^{10}}{2^9} + \frac{3^{10}}{2^9} - 3$ Combine the terms with the same base and exponent: $S = 2 \cdot \frac{3^{10}}{2^9} - 3$ Since $2 = \frac{2^1}{1}$, we can simplify the first term: $S = \frac{2^1 \cdot 3^{10}}{2^9} - 3$ $S = \frac{3^{10}}{2^{9-1}} - 3$ $S = \frac{3^{10}}{2^8} - 3$ We need to express this in the form $\alpha\left(\frac{3}{2}\right)^9-\beta$. Let's manipulate $\frac{3^{10}}{2^8}$: $\frac{3^{10}}{2^8} = \frac{3^1 \cdot 3^9}{2^8}$ To get $\left(\frac{3}{2}\right)^9$, we need $2^9$ in the denominator. We can multiply the numerator and denominator by 2: $\frac{3 \cdot 3^9}{2^8} = \frac{3 \cdot 3^9 \cdot 2}{2^8 \cdot 2} = \frac{6 \cdot 3^9}{2^9} = 6 \left(\frac{3}{2}\right)^9$ So, the entire sum $S$ becomes: $S = 6 \left(\frac{3}{2}\right)^9 - 3$ Comparing this with the given form $\alpha\left(\frac{3}{2}\right)^9-\beta$, we can identify $\alpha$ and $\beta$: $\alpha = 6$ $\beta = 3$ As specified, $\alpha, \beta \in \mathbb{N}$ (natural numbers), which holds true for $6$ and $3$.

6. Final Calculation

The problem asks for the value of $(\alpha+\beta)^2$. $(\alpha+\beta)^2 = (6+3)^2$ $(\alpha+\beta)^2 = (9)^2$ $(\alpha+\beta)^2 = 81$

7. Tips for Success and Common Pitfalls

• Decomposition: When the numerator of a term within a summation is a sum (like $r+3$), always consider splitting the summation into separate parts. This often simplifies each part significantly.
• Binomial Coefficient Properties: Memorize and understand the common properties of binomial coefficients, especially $r \cdot {}^n C_r = n \cdot {}^{n-1} C_{r-1}$ and $\frac{{}^n C_r}{r+1} = \frac{{}^{n+1} C_{r+1}}{n+1}$. They are key to solving many binomial summation problems.
• Adjusting Limits: Pay close attention to the starting and ending limits of the summation. If a sum starts from $r=1$ but a standard binomial expansion starts from $r=0$, remember to subtract the missing term (the $r=0$ term).
• Algebraic Manipulation: Be careful with exponents and fractions during simplification. The goal is often to match a specific target form, so work backwards from the target if necessary to guide your simplification steps.

8. Summary/Key Takeaway

This problem demonstrates a classic strategy for evaluating complex binomial summations:

1. Split the sum based on the numerator structure.
2. Apply binomial coefficient properties to simplify terms like $r \cdot {}^n C_r$.
3. Adjust summation limits and constant factors to match the standard binomial expansion $(1+x)^n$.
4. Combine and simplify the resulting expressions to reach the desired form. The ability to recognize and apply these techniques efficiently is crucial for success in JEE Mathematics.