Prove by induction on, that $${p_n} = A{\alpha ^n} + B{\beta ^n}$$ for all $$n \ge 1,$$ where $$\alpha $$ and $$\beta $$ are the roots of quadratic equation $${x^2} - \left( {1 - p} \right)x - p\left( {1 - p} \right) = 0$$ and $$A = {{{p^2} + \beta - 1} \over {\alpha \beta - {\alpha ^2}}},B = {{{p^2} + \alpha - 1} \over {\alpha \beta - {\beta ^2}}}.$$
$${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$$ for all $$n = 1,2,....$$ (Here, 'well-defined' means that the denominator in the expression for $${\alpha _{n + 1}}$$ is not zero.)
$$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$$
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 $$x$$.
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 $$\left( {\matrix{ n \cr m \cr } } \right) + 2\left( {\matrix{ {n - 1} \cr m \cr } } \right) + 3\left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {n - m + 1} \right)\left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 2} \cr {m + 2} \cr } } \right).$$.
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