Question

Let $\alpha = \sum_{r=0}^n (4r^2 + 2r + 1) \binom{n}{r}$ and $\beta = \left( \sum_{r=0}^n \frac{\binom{n}{r}}{r+1} \right) + \frac{1}{n+1}$. If $140 < \frac{2\alpha}{\beta} < 281$, then the value of $n$ is _____.

Answer 1

Correct Answer: 5

\[ \alpha = \sum_{r=0}^{n} (4r^2 + 2r + 1) \cdot \binom{n}{r} \] \[ \alpha = 4 \sum_{r=0}^{n} r^2 \cdot \binom{n-1}{r-1} + 2 \sum_{r=0}^{n} r \cdot \binom{n-1}{r} + \sum_{r=0}^{n} n \cdot \binom{n}{r} \] \[ + 4n \sum_{r=0}^{n} (n-1) \cdot \binom{n-1}{r-1} + 2n \sum_{r=0}^{n} \binom{n-1}{r-1} + \sum_{r=0}^{n} n \cdot \binom{n}{r} \] \[ \alpha = 4n(n-1) \cdot 2n^2 + 4n \cdot n^2 + 2n \cdot 2n \cdot n^2 \] \[ \alpha = 2n(n+1)^2 \] \[ \beta = \sum_{r=0}^{n} \binom{r+1}{r+1} + \frac{1}{n+1} \] \[ \beta = \sum_{r=0}^{n} \binom{n+1}{r+1} + \frac{1}{n+1} \] \[ \beta = \frac{1}{n+1} \left( 1 + n+1 \cdot \binom{C1} + ... + \binom{n+1}{n+1} \right) \] \[ \beta = \frac{2^{n+1}}{n+1} \] \[ \frac{2a}{\beta} = \frac{2^{n+1} (n+1)^2}{2^{n+1}} \cdot (n+1)^3 = (n+1)^3 \] \[ 140 < (n+1)^3 < 281 \] \[ n = 4 \Rightarrow (n+1)^3 = 125 \] \[ n = 5 \Rightarrow (n+1)^3 = 216 \] \[ n = 6 \Rightarrow (n+1)^3 = 343 \] \[ \Rightarrow n = 5 \] 

Answer 2

The expression for \(\alpha\) is:

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

Expand the summation:

$$\alpha = \sum_{r=0}^{n} 4r^2 \binom{n}{r} + \sum_{r=0}^{n} 2r \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r}.$$

Using standard summation identities for binomial coefficients:

$$\sum_{r=0}^{n} r \binom{n}{r} = n \cdot 2^{n-1}, \quad \sum_{r=0}^{n} r^2 \binom{n}{r} = n(n-1) \cdot 2^{n-2}, \quad \sum_{r=0}^{n} \binom{n}{r} = 2^n.$$

Substitute these results:

$$\alpha = 4n(n-1) \times 2^{n-2} + 2n \times 2^{n-1} + 2^n.$$

Factorize:

$$\alpha = 2^{n-2} [4n(n-1) + 8n + 4].$$

Simplify:

$$\alpha = 2^{n-2} \times 2n(n+1) = 2n(n+1)2^{n-2}.$$

Now for \(\beta\):

$$\beta = \sum_{r=0}^{n} \binom{n}{r+1} + \frac{1}{n+1}.$$

Rewrite the summation:

$$\sum_{r=0}^{n} \binom{n}{r+1} = \sum_{r=1}^{n} \binom{n}{r} = \sum_{r=0}^{n} \binom{n}{r} - \binom{n}{0}.$$

Using the summation of binomial coefficients:

$$\sum_{r=0}^{n} \binom{n}{r} = 2^n, \quad \binom{n}{0} = 1.$$

Thus:

$$\sum_{r=0}^{n} \binom{n}{r+1} = 2^n - 1.$$

Therefore:

$$\beta = (2^n - 1) + \frac{1}{n+1}.$$

Now calculate \(\frac{2\alpha}{\beta}\):

$$\frac{2\alpha}{\beta} = \frac{2 \times 2^{n-2} \times 2n(n+1)}{(2^n - 1) + \frac{1}{n+1}}.$$

Simplify:

$$\frac{2\alpha}{\beta} = \frac{2^{n-1} \times n(n+1)^2}{2^n - 1 + \frac{1}{n+1}}.$$

Testing values of \(n\), find \(140 < \frac{2\alpha}{\beta} < 281\):

For \(n = 4\):

$$\frac{2\alpha}{\beta} = \frac{2 \times 4 \times 5^2}{2^5 - 1 + \frac{1}{5}} = \frac{200}{31.2} \approx 125 \quad \text{(Too low)}.$$

For \(n = 5\):

$$\frac{2\alpha}{\beta} = \frac{2 \times 5 \times 6^2}{2^6 - 1 + \frac{1}{6}} = \frac{360}{63.17} \approx 216.$$

For \(n = 6\):

$$\frac{2\alpha}{\beta} = \frac{2 \times 6 \times 7^2}{2^7 - 1 + \frac{1}{6}}.$$