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

Using standard binomial identities:

$$\alpha = \sum_{r=0}^n (4r^2+2r+1)\binom{n}{r} = 4[n(n+1)2^{n-2}] + 2[n2^{n-1}] + 2^n = 2^n(n+1)^2$$

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

Divide the two expressions:

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

Given $$140 < (n+1)^3 < 281$$, the only matching perfect cube is $$6^3 = 216$$, so $$n+1 = 6 \implies n = 5$$.