Let `alpha = sum_(r = 0)^n (4r^2 + 2r + 1)^nC_r` and`beta = (sum_(r = 0)^n frac{"^nC_r}{r+1})+1/(n+1)`. If `140 < frac{2 alpha}{beta} < 281`, then the value of `n` is _____________.

Answer

To solve the problem, we need to evaluate the expressions for \(\alpha\) and \(\beta\) and then find the value of \(n\) such that \(140 < \frac{2\alpha}{\beta} < 281\).

### Step 1: Evaluate \(\alpha\)

\[ \alpha = \sum_{r=0}^{n} (4r^2 + 2r + 1) \binom{n}{r} \]

We can break this sum into three parts:

\[ \alpha = \sum_{r=0}^{n} 4r^2 \binom{n}{r} + \sum_{r=0}^{n} 2r \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r} \]

Using known formulas:

1. \(\sum_{r=0}^{n} r \binom{n}{r} = n \cdot 2^{n-1}\)
2. \(\sum_{r=0}^{n} r^2 \binom{n}{r} = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}\)
3. \(\sum_{r=0}^{n} \binom{n}{r} = 2^n\)

Now substituting these into our expression for \(\alpha\):

\[ \sum_{r=0}^{n} 4r^2 \binom{n}{r} = 4\left(n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}\right) \]

\[ \sum_{r=0}^{n} 2r \binom{n}{r} = 2(n \cdot 2^{n-1}) \]

\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

Combining these results:

\[ \alpha = 4\left(n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}\right) + 2(n \cdot 2^{n-1}) + 2^n \]

\[ = 4n(n-1) \cdot 2^{n-2} + 4n \cdot 2^{n-1} + 2n \cdot 2^{n-1} + 2^n \]

\[ = 4n(n-1) \cdot 2^{n-2} + 6n \cdot 2^{n-1} + 2^n \]

### Step 2: Evaluate \(\beta\)

\[ \beta = \left(\sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1}\right) + \frac{1}{n+1} \]

Using the known result:

\[ \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} = \frac{1}{n+1} \cdot 2^{n+1} \]

Thus,

\[ \beta = \frac{1}{n+1} \cdot 2^{n+1} + \frac{1}{n+1} \]

\[ = \frac{2^{n+1} + 1}{n+1} \]

### Step 3: Calculate \(\frac{2\alpha}{\beta}\)

Now we can compute \(\frac{2\alpha}{\beta}\):

\[ \frac{2\alpha}{\beta} = \frac{2 \left(4n(n-1) \cdot 2^{n-2} + 6n \cdot 2^{n-1} + 2^n\right)}{\frac{2^{n+1} + 1}{n+1}} \]

This simplifies to:

\[ = \frac{2(n+1) \left(4n(n-1) \cdot 2^{n-2} + 6n \cdot 2^{n-1} + 2^n\right)}{2^{n+1} + 1} \]

### Step 4: Solve the inequality

We need to solve:

\[ 140 < \frac{2\alpha}{\beta} < 281 \]

This involves substituting values for \(n\) and checking the inequalities.

### Step 5: Finding \(n\)

After calculating for \(n = 5\) and \(n = 6\):

1. For \(n = 5\):
- Calculate \(\frac{2\alpha}{\beta}\) and check if it lies within the bounds.

2. For \(n = 6\):
- Repeat the calculation and check.

After evaluating, we find that the value of \(n\) that satisfies the inequality is:

\[ \boxed{5} \]