Use mathematical induction to show that
$${\left( {25} \right)^{n + 1}} - 24n + 5735$$ is divisible by $${\left( {24} \right)^2}$$ for all $$ = n = 1,2,...$$

For any positive integer $$m$$, $$n$$ (with $$n \ge m$$), let $$\left( {\matrix{ n \cr m \cr } } \right) = {}^n{C_m}$$
Prove that $$\left( {\matrix{ n \cr m \cr } } \right) + \left( {\matrix{ {n - 1} \cr m \cr } } \right) + \left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 1} \cr {m + 2} \cr } } \right)$$

Hence or otherwise, prove that $$\left( {\matrix{ n \cr m \cr } } \right) + 2\left( {\matrix{ {n - 1} \cr m \cr } } \right) + 3\left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {n - m + 1} \right)\left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 2} \cr {m + 2} \cr } } \right).$$.

For every possitive integer $$n$$, prove that
$$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$$
Hence or otherwise, prove that $$\left[ {\sqrt n + \sqrt {\left( {n + 1} \right)} } \right] = \left[ {\sqrt {4n + 1} \,\,} \right],$$
where $$\left[ x \right]$$ denotes the gratest integer not exceeding $$x$$.

Let $$a,\,b,\,c$$ be possitive real numbers such that $${b^2} - 4ac > 0$$ and let $${\alpha _1} = c.$$ Prove by induction that $${\alpha _{n + 1}} = {{a\alpha _n^2} \over {\left( {{b^2} - 2a\left( {{\alpha _1} + {\alpha _2} + ... + {\alpha _n}} \right)} \right)}}$$ is well-defined and
$${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$$ for all $$n = 1,2,....$$ (Here, 'well-defined' means that the denominator in the expression for $${\alpha _{n + 1}}$$ is not zero.)