1
IIT-JEE 2000
Subjective
+6
-0
For any positive integer $$m$$, $$n$$ (with $$n \ge m$$), let $$\left( {\matrix{
n \cr
m \cr
} } \right) = {}^n{C_m}$$
Prove that $$\left( {\matrix{ n \cr m \cr } } \right) + \left( {\matrix{ {n - 1} \cr m \cr } } \right) + \left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 1} \cr {m + 2} \cr } } \right)$$
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 $$\left( {\matrix{ n \cr m \cr } } \right) + 2\left( {\matrix{ {n - 1} \cr m \cr } } \right) + 3\left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {n - m + 1} \right)\left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 2} \cr {m + 2} \cr } } \right).$$.
2
IIT-JEE 2000
Subjective
+6
-0
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 $$x$$.
$$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$$
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 $$x$$.
3
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 $$A = {{{p^2} + \beta - 1} \over {\alpha \beta - {\alpha ^2}}},B = {{{p^2} + \alpha - 1} \over {\alpha \beta - {\beta ^2}}}.$$
4
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.)
$${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$$ for all $$n = 1,2,....$$ (Here, 'well-defined' means that the denominator in the expression for $${\alpha _{n + 1}}$$ is not zero.)
Questions Asked from Mathematical Induction and Binomial Theorem (Subjective)
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