Quadratic Binomial Fusion

$\displaystyle 1 + \sum_{r = 1}^{10} \left [ 3^r \binom{10}{r} + r \binom{10}{r} \right ] = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$\displaystyle 1 + \sum_{r = 1}^{10} \binom{10}{r} 3^r + \sum_{r = 1}^{10} \binom{10}{r} r = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$\displaystyle 1 + \sum_{r = 0}^{10} \binom{10}{r} 3^r - \binom{10}{0} 3^0 + \sum_{r = 1}^{10} \binom{10}{r} r = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$1 +(1 + 3)^{10} - 1 + 5 \times 2^{10} = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$4^{10} + 5 \times 2^{10} = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$2^{10} \times 2^{10} + 5 \times 2^{10} = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$2^{10} \left( 2^{10} + 5 \right) = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

$2^{10} \left( 4^{5} + 5 \right) = 2^{10} \left (\alpha \cdot 4^5 + \beta \right )$

We have

$\alpha = 1$ and $\beta = 5$

$f(x) = x^2 - 2x + 1 -k^2$ . Let the roots of $f(x) = 0$ are $x_{1}$ and $x_{2}$ , where $x_{1}$ < $x_{2}$ , we have

$x_{1} + x_{2} = 2$ and $x_{1} \times x_{2} = 1 - k^2$

Since $\alpha, \beta$ lies strictly in between the roots of $f(x) = 0$ and to find the smallest positive integral value of $k$ :

$x_{1} = 0$ or $x_{2} = 6$

For $x_{1} = 0$ we have $x_{2} = 2$ , but $\beta$ doesn't lie on that range

For $x_{2} = 6$ we have $x_{1} = -4$ then

$-4 \times 6 = 1 - k^2$

$-24 = 1 - k^2$

$k^2 = 25$

$\boxed{k = 5}$

Reduce the given expression in required form to get alpha and beta as 1 and 5 respectively.

Now simply use location of roots concept.

I wish to ask @Krishna Sharma , is there a rigorious way to show that (i mean other than comparing terms) that there is no other integral solution to (alpha) and (beta) other than 1 and 5