1. Compute $\alpha$

Given

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

we split it as

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

Now use standard identities:

• $\sum_{r=0}^n \binom nr = 2^n$
• $\sum_{r=0}^n r\binom nr = n2^{n-1}$
• $\sum_{r=0}^n r(r-1)\binom nr = n(n-1)2^{n-2}$

Hence

$$\sum_{r=0}^n r^2\binom nr =\sum r(r-1)\binom nr + \sum r\binom nr =n(n-1)2^{n-2}+n2^{n-1}.$$

So

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

Therefore,

$$\alpha=4\cdot 2^{n-2}n(n+1)+2\cdot n2^{n-1}+2^n.$$

Simplify:

$$4\cdot 2^{n-2}=2^n,$$

so

$$\alpha=2^n n(n+1)+n2^n+2^n=2^n(n^2+2n+1)=2^n(n+1)^2.$$

Thus,

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

---

2. Compute $\beta$

Given

$$\beta=\left(\sum_{r=0}^n \frac{\binom nr}{r+1}\right)+\frac{1}{n+1}.$$

Use the identity

$$\frac{1}{r+1}\binom nr = \frac{1}{n+1}\binom{n+1}{r+1}.$$

Hence

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

Put $k=r+1$. Then $k=1$ to $n+1$:

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

So

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

Therefore

\beta=\frac{2^{n+1}-1}{n+1}+\frac{1}{n+1}=rac{2^{n+1}}{n+1}.

Thus,

$$\boxed{\beta=\frac{2^{n+1}}{n+1}}.$$

---

3. Find $\dfrac{2\alpha}{\beta}$

Substitute the expressions for $\alpha$ and $\beta$:

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

Since $2\cdot 2^n=2^{n+1}$,

$$\frac{2\alpha}{\beta}=\frac{2^{n+1}(n+1)^2}{2^{n+1}/(n+1)}=(n+1)^3.$$

So the inequality becomes

$$140<(n+1)^3<281.$$

---

4. Solve the inequality

Check nearby cubes:

$$5^3=125,\qquad 6^3=216,\qquad 7^3=343.$$

Thus,

$$140<(n+1)^3<281$$

implies

$$(n+1)^3=216 \Rightarrow n+1=6 \Rightarrow n=5.$$

Hence,

$$\boxed{n=5}.$$

---

5. Compare with stored answer

Stored correct answer: $5$

Our derived answer is also $5$, so they agree.