Prove that there exists no complex number z such that $$\left| z \right| < {1 \over 3}\,and\,\sum\limits_{r = 1}^n {{a_r}{z^r}} = 1$$ where $$\left| {{a_r}} \right| < 2$$.

If $${x^2} + \left( {a - b} \right)x + \left( {1 - a - b} \right) = 0$$ where $$a,\,b\, \in \,R$$ then find the values of a for which equation has unequal real roots for all values of $$b$$.

Prove that
$${2^k}\left( {\matrix{ n \cr 0 \cr } } \right)\left( {\matrix{ n \cr k \cr } } \right) - {2^{^{k - 1}\left( {\matrix{ n \cr 2 \cr } } \right)}}\left( {\matrix{ n \cr 1 \cr } } \right)\left( {\matrix{ {n - 1} \cr {k - 1} \cr } } \right)$$
$$ + {2^{k - 2}}\left( {\matrix{ {n - 2} \cr {k - 2} \cr } } \right) - .....{\left( { - 1} \right)^k}\left( {\matrix{ n \cr k \cr } } \right)\left( {\matrix{ {n - k} \cr 0 \cr } } \right) = {\left( {\matrix{ n \cr k \cr } } \right)^ \cdot }$$

If a, b, c are in A.P., $${a^2}$$, $${b^2}$$, $${c^2}$$ are in H.P., then prove that either a = b = c or a, b, $${ - {c \over 2}}$$ form a G.P.