Given that $${C_1} + 2{C_2}x + 3{C_3}{x^2} + ......... + 2n{C_{2n}}{x^{2n - 1}} = 2n{\left( {1 + x} \right)^{2n - 1}}$$
where $${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$$
Prove that $${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$$

$${}^n{C_{r - 1}} = 36,{}^n{C_r} = 84\,\,and\,\,{}^n{C_{r + 1}} = 126$$, then r is :

The harmonic mean of two numbers is 4.Their arithmetic mean $$A$$ and the geometric mean $$G$$ satisfy the relation. $$2A + {G^2} = 27$$

The points $$\left( { - a,\, - b} \right),\,\left( {0,\,0} \right),\,\left( {a,\,b} \right)$$ and $$\left( {{a^2},\,ab} \right)$$ are :