Mathematical Induction and Binomial Theorem · Mathematics · JEE Advanced
Numerical
where $${}^{10}{C_r}$$, r $$ \in $${1, 2, ..., 10} denote binomial coefficients. Then, the value of $${1 \over {1430}}X$$ is ..........
MCQ (More than One Correct Answer)
MCQ (Single Correct Answer)
Then $$\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)} $$ is equal to
is where $$\left( {\matrix{ n \cr r \cr } } \right) = {}^n{C_r}$$
where $$n$$ is an even positive integer, is equal to
Subjective
$${2^k}\left( {\matrix{ n \cr 0 \cr } } \right)\left( {\matrix{ n \cr k \cr } } \right) - {2^{^{k - 1}\left( {\matrix{ n \cr 2 \cr } } \right)}}\left( {\matrix{ n \cr 1 \cr } } \right)\left( {\matrix{ {n - 1} \cr {k - 1} \cr } } \right)$$
$$ + {2^{k - 2}}\left( {\matrix{ {n - 2} \cr {k - 2} \cr } } \right) - .....{\left( { - 1} \right)^k}\left( {\matrix{ n \cr k \cr } } \right)\left( {\matrix{ {n - k} \cr 0 \cr } } \right) = {\left( {\matrix{ n \cr k \cr } } \right)^ \cdot }$$
$${\left( {25} \right)^{n + 1}} - 24n + 5735$$ is divisible by $${\left( {24} \right)^2}$$ for all $$ = n = 1,2,...$$
$$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$$
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 $$\left( {\matrix{ n \cr m \cr } } \right) + 2\left( {\matrix{ {n - 1} \cr m \cr } } \right) + 3\left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {n - m + 1} \right)\left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 2} \cr {m + 2} \cr } } \right).$$.
$${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$$ for all $$n = 1,2,....$$ (Here, 'well-defined' means that the denominator in the expression for $${\alpha _{n + 1}}$$ is not zero.)
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 $$A = {{{p^2} + \beta - 1} \over {\alpha \beta - {\alpha ^2}}},B = {{{p^2} + \alpha - 1} \over {\alpha \beta - {\beta ^2}}}.$$
for each non-be gatuve integer $$m \le n.$$ $$\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$$
[Hint: You may use the fact that $${\left( {1 + x} \right)^{\left( {m + 1} \right)p}} = {\left( {1 + x} \right)^p}{\left( {1 + x} \right)^{mp}}$$]
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 $$0 \le x,y \le \pi .$$}
Show that $$a_0^2 - a_1^2 + a_2^2...... + {a_{2n}}{}^2 = {a_n}$$
$${\tan ^{ - 1}}\left( {1/3} \right) + {\tan ^{ - 1}}\left( {1/7} \right) + ........{\tan ^{ - 1}}\left\{ {1/\left( {{n^2} + n + 1} \right)} \right\} = {\tan ^{ - 1}}\left\{ {n/\left( {n + 2} \right)} \right\}$$
(i) is an integer and (ii) is not divisible by $$p$$
$$\sum\limits_{m = 0}^k {\left( {n - m} \right)} {{\left( {r + m} \right)!} \over {m!}} = {{\left( {r + k + 1} \right)!} \over {k!}}\left[ {{n \over {r + 1}} - {k \over {r + 2}}} \right]$$
$${C_0} - {2^2}{C_1} + {3^2}{C_2}\,\, - \,..... + {\left( { - 1} \right)^n}{\left( {n + 1} \right)^2}{C_n} = 0,\,\,\,\,n > 2,\,\,$$ where $${C_r} = {}^n{C_r}.$$
where $$m,\,n,\,k$$ are positive integers, and $${}^p{C_q} = 0$$ for $$p < q.$$
$${S_n} = 1 + {{q + 1} \over 2} + {\left( {{{q + 1} \over 2}} \right)^2} + ........ + {\left( {{{q + 1} \over 2}} \right)^n}\,\,\,,q \ne 1$$
Prove that $${}^{n + 1}{C_1} + {}^{n + 1}{C_2}{s_1} + {}^{n + 1}{C_3}{s_2} + ..... + {}^{n + 1}{C_n}{s_n} = {2^n}{S_n}$$
where $${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$$
Prove that $${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$$