1
IIT-JEE 2000
Subjective
+6
-0
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.)
2
IIT-JEE 2000
Subjective
+5
-0
A coin probability $$p$$ of showing head when tossed. It is tossed $$n$$ times. Let $${p_n}$$ denote the probability that no two (or more) consecutive heads occur. Prove that $${p_1} = 1,\,\,{p_2} = 1 - {p^2}$$ and $${p_n} = \left( {1 - p} \right).\,\,{p_{n - 1}} + p\left( {1 - p} \right){p_{n - 2}}$$ for all $$n \ge 3.$$

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 $$A = {{{p^2} + \beta - 1} \over {\alpha \beta - {\alpha ^2}}},B = {{{p^2} + \alpha - 1} \over {\alpha \beta - {\beta ^2}}}.$$

3
IIT-JEE 1999
Subjective
+10
-0
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 $$m \le n.$$ $$\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$$

4
IIT-JEE 1998
Subjective
+8
-0
Let $$p$$ be a prime and $$m$$ a positive integer. By mathematical induction on $$m$$, or otherwise, prove that whenever $$r$$ is an integer such that $$p$$ does not divide $$r$$, $$p$$ divides $${}^{np}{C_r},$$

[Hint: You may use the fact that $${\left( {1 + x} \right)^{\left( {m + 1} \right)p}} = {\left( {1 + x} \right)^p}{\left( {1 + x} \right)^{mp}}$$]

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