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 $$m \le n.$$ $$\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$$

Let $${a_1},{a_2},......{a_{10}}$$ be in $$A,\,P,$$ and $${h_1},{h_2},......{h_{10}}$$ be in H.P. If $${a_1} = {h_1} = 2$$ and $${a_{10}} = {h_{10}} = 3,$$ then $${a_4}{h_7}$$ is

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

Let a, b, c, d be real numbers in G.P. If u, v, w, satisfy the system of equations
u + 2v + 3w = 6
4u + 5v + 6w = 12
6u + 9v = 4

then show that the roots of the equation $$\left( {{1 \over u} + {1 \over v} + {1 \over w}} \right){x^2}$$
$$ + [{(b - c)^2} + {(c - a)^2} + {(d - b)^2}]x + u + v + w = 0$$ and $$20{x^2} + 10{(a - d)^2}x - 9 = 0$$ are reciprocals of each other.