1

IIT-JEE 2000

Subjective
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 $$\left( {\matrix{ n \cr m \cr } } \right) + 2\left( {\matrix{ {n - 1} \cr m \cr } } \right) + 3\left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {n - m + 1} \right)\left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 2} \cr {m + 2} \cr } } \right).$$.

Answer

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2

IIT-JEE 1999

Subjective
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 $$m \le n.$$ $$\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$$

Answer

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3

IIT-JEE 1998

Subjective
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 $${\left( {1 + x} \right)^{\left( {m + 1} \right)p}} = {\left( {1 + x} \right)^p}{\left( {1 + x} \right)^{mp}}$$]

Answer

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4

IIT-JEE 1997

Subjective
Let $$0 < {A_i} < n$$ for $$i = 1,\,2....,\,n.$$ Use mathematical induction to prove that $$$\sin {A_1} + \sin {A_2}....... + \sin {A_n} \le n\,\sin \,\,\left( {{{{A_1} + {A_2} + ...... + {A_n}} \over n}} \right)$$$

where $$ \ge 1$$ is a natural number. {You may use the fact that $$p\sin x + \left( {1 - p} \right)\sin y \le \sin \left[ {px + \left( {1 - p} \right)y} \right],$$ where $$0 \le p \le 1$$ and $$0 \le x,y \le \pi .$$}

Answer

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