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

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 $n = 6$:

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

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 $n = 6$:

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