Let $$n$$ be any positive integer. Prove that $$$\sum\limits_{k = 0}^m {{{\left( {\matrix{ {2n - k} \cr k \cr } } \right)} \over {\left( {\matrix{ {2n - k} \cr n \cr } } \right)}}.{{\left( {2n - 4k + 1} \right)} \over {\left( {2n - 2k + 1} \right)}}{2^{n - 2k}} = {{\left( {\matrix{ n \cr m \cr } } \right)} \over {\left( {\matrix{ {2n - 2m} \cr {n - m} \cr } } \right)}}{2^{n - 2m}}} $$$

for each non-be gatuve integer $$m \le n.$$ $$\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$$

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the coefficients of $$x$$ and $${x^2}$$ are $$3$$ and $$-6$$ respectively, then $$m$$ is

If $${x_1},\,{x_2},\,{x_3}$$ as well as $${y_1},\,{y_2},\,{y_3}$$, are in G.P. with the same common ratio, then the points $$\left( {{x_1},\,{y_1}} \right),\left( {{x_2},\,{y_2}} \right)$$ and $$\left( {{x_3},\,{y_3}} \right).$$

The harmonic mean of the roots of the equation $$\left( {5 + \sqrt 2 } \right){x^2} - \left( {4 + \sqrt 5 } \right)x + 8 + 2\sqrt 5 = 0$$ is