Question

If $ \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \alpha \left( \frac{3}{2} \right)^9 - \beta $, $ \alpha, \beta \in \mathbb{N} $, then $ (\alpha + \beta)^2 $ is equal to

• A. 27
• B. 9
• C. 81
• D. 18

Hint:

Use the identity \( r \cdot {^nC_r} = n \cdot {^{n-1}C_{r-1}} \) to simplify the summation. Also, remember the binomial expansion \( (1+x)^n = \sum_{r=0}^{n} {^nC_r} x^r \).

Answer 1

The Correct Option is C

Given:
\[ \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \alpha \left( \frac{3}{2} \right)^9 - \beta,\quad \alpha, \beta \in \mathbb{N} \] We split the sum as: \[ \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} = \sum_{r=1}^{9} \left( \frac{r}{2^r} \cdot {^9C_r} \right) + \sum_{r=1}^{9} \left( \frac{3}{2^r} \cdot {^9C_r} \right) \] Now, use the identity: \[ \frac{r}{2^r} \cdot {^9C_r} = \frac{9}{2^r} \cdot {^8C_{r-1}} \] So, \[ \sum_{r=1}^{9} \frac{r}{2^r} \cdot {^9C_r} = 9 \sum_{r=1}^{9} \frac{1}{2^r} \cdot {^8C_{r-1}} = \frac{9}{2} \sum_{r=1}^{9} {^8C_{r-1}} \cdot \left( \frac{1}{2} \right)^{r-1} \] Make the substitution \( s = r-1 \Rightarrow s = 0 \) to \( 8 \): \[ = \frac{9}{2} \sum_{s=0}^{8} {^8C_s} \left( \frac{1}{2} \right)^s = \frac{9}{2} \cdot \left(1 + \frac{1}{2}\right)^8 = \frac{9}{2} \cdot \left( \frac{3}{2} \right)^8 \] Now for the second term: \[ \sum_{r=1}^{9} \frac{3}{2^r} \cdot {^9C_r} = 3 \left( \sum_{r=0}^{9} {^9C_r} \cdot \left( \frac{1}{2} \right)^r - {^9C_0} \cdot 1 \right) = 3 \left( \left( 1 + \frac{1}{2} \right)^9 - 1 \right) = 3 \left( \left( \frac{3}{2} \right)^9 - 1 \right) \] Adding both parts: \[ \frac{9}{2} \cdot \left( \frac{3}{2} \right)^8 + 3 \left( \left( \frac{3}{2} \right)^9 - 1 \right) = \left( \frac{9}{2} \cdot \frac{2}{3} + 3 \right) \cdot \left( \frac{3}{2} \right)^9 - 3 = (3 + 3) \cdot \left( \frac{3}{2} \right)^9 - 3 = 6 \left( \frac{3}{2} \right)^9 - 3 \]
Thus, \( \alpha = 6 \), \( \beta = 3 \) \[ \therefore (\alpha + \beta)^2 = (6 + 3)^2 = 81 \]

Answer 2

The problem asks to evaluate the summation \( \sum_{r=1}^{9} \left( \frac{r+3}{2^r} \right) \cdot {^9C_r} \) and equate it to the form \( \alpha \left( \frac{3}{2} \right)^9 - \beta \) to find the value of \( (\alpha + \beta)^2 \), where \( \alpha \) and \( \beta \) are natural numbers.

Concept Used:

This problem utilizes properties of binomial coefficients and the binomial theorem. The key formulas are:

1. Binomial Theorem: The expansion of \( (1+x)^n \) is given by:

\[ (1+x)^n = \sum_{r=0}^{n} {^nC_r} x^r = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + \ldots + {^nC_n}x^n \]

2. Binomial Coefficient Identity: A useful property for summations involving \( r \cdot {^nC_r} \) is:

\[ r \cdot {^nC_r} = n \cdot {^{n-1}C_{r-1}} \]

Step-by-Step Solution:

Step 1: Let \( S \) be the given summation. We can split the term \( \frac{r+3}{2^r} \) into two parts and separate the summation accordingly.

\[ S = \sum_{r=1}^{9} \left( \frac{r}{2^r} + \frac{3}{2^r} \right) {^9C_r} \] \[ S = \sum_{r=1}^{9} \frac{r}{2^r} {^9C_r} + \sum_{r=1}^{9} \frac{3}{2^r} {^9C_r} \]

Let's denote these two summations as \( S_1 \) and \( S_2 \).

\[ S_1 = \sum_{r=1}^{9} r \frac{{^9C_r}}{2^r}, \quad S_2 = \sum_{r=1}^{9} 3 \frac{{^9C_r}}{2^r} \]

Step 2: Evaluate the first summation, \( S_1 \). We use the identity \( r \cdot {^9C_r} = 9 \cdot {^8C_{r-1}} \).

\[ S_1 = \sum_{r=1}^{9} \frac{9 \cdot {^8C_{r-1}}}{2^r} \]

Let's factor out the constant 9 and adjust the power of 2:

\[ S_1 = 9 \sum_{r=1}^{9} \frac{{^8C_{r-1}}}{2 \cdot 2^{r-1}} = \frac{9}{2} \sum_{r=1}^{9} {^8C_{r-1}} \left(\frac{1}{2}\right)^{r-1} \]

Now, let \( k = r-1 \). As \( r \) goes from 1 to 9, \( k \) goes from 0 to 8.

\[ S_1 = \frac{9}{2} \sum_{k=0}^{8} {^8C_k} \left(\frac{1}{2}\right)^k \]

The summation is the binomial expansion of \( \left(1 + \frac{1}{2}\right)^8 \).

\[ S_1 = \frac{9}{2} \left(1 + \frac{1}{2}\right)^8 = \frac{9}{2} \left(\frac{3}{2}\right)^8 = \frac{3^2}{2} \cdot \frac{3^8}{2^8} = \frac{3^{10}}{2^9} \]

We can rewrite this as:

\[ S_1 = 3 \cdot \frac{3^9}{2^9} = 3 \left(\frac{3}{2}\right)^9 \]

Step 3: Evaluate the second summation, \( S_2 \).

\[ S_2 = 3 \sum_{r=1}^{9} {^9C_r} \left(\frac{1}{2}\right)^r \]

The summation resembles the binomial expansion of \( \left(1 + \frac{1}{2}\right)^9 \), but it starts from \( r=1 \) instead of \( r=0 \).

We know that \( \sum_{r=0}^{9} {^9C_r} \left(\frac{1}{2}\right)^r = \left(1 + \frac{1}{2}\right)^9 = \left(\frac{3}{2}\right)^9 \).

So, we can write:

\[ \sum_{r=1}^{9} {^9C_r} \left(\frac{1}{2}\right)^r = \left( \sum_{r=0}^{9} {^9C_r} \left(\frac{1}{2}\right)^r \right) - {^9C_0} \left(\frac{1}{2}\right)^0 = \left(\frac{3}{2}\right)^9 - 1 \]

Substituting this back into the expression for \( S_2 \):

\[ S_2 = 3 \left[ \left(\frac{3}{2}\right)^9 - 1 \right] = 3 \left(\frac{3}{2}\right)^9 - 3 \]

Step 4: Combine \( S_1 \) and \( S_2 \) to find the total sum \( S \).

\[ S = S_1 + S_2 = 3 \left(\frac{3}{2}\right)^9 + \left( 3 \left(\frac{3}{2}\right)^9 - 3 \right) \] \[ S = 6 \left(\frac{3}{2}\right)^9 - 3 \]

Final Computation & Result:

We are given that the sum is equal to \( \alpha \left( \frac{3}{2} \right)^9 - \beta \).

Comparing our result with the given form:

\[ 6 \left(\frac{3}{2}\right)^9 - 3 = \alpha \left( \frac{3}{2} \right)^9 - \beta \]

We can see that \( \alpha = 6 \) and \( \beta = 3 \). Both are natural numbers, as required.

Now we need to find the value of \( (\alpha + \beta)^2 \).

\[ \alpha + \beta = 6 + 3 = 9 \] \[ (\alpha + \beta)^2 = 9^2 = 81 \]

The value of \( (\alpha + \beta)^2 \) is 81.