Suppose $\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $. Then the value of $\alpha$ is ___________
Explanation:
Concept :
(1) ${}^n{C_r} = {n \over r}\,.\,{}^{n - 1}{C_{r - 1}}$
Given,
$\sum\limits_{r = 0}^{2023} {{r^2}\,.\,{}^{2023}{C_r}} $
$ = \sum\limits_{r = 0}^{2023} {{r^2}\,.\,{{2023} \over r}\,.{}^{2022}{C_{r - 1}}} $
$ = 2023\sum\limits_{r = 0}^{2023} {{r}\,.\,{}^{2022}{C_{r - 1}}} $
$ = 2023\sum\limits_{r = 0}^{2023} {[(r - 1) + 1]\,.\,{}^{2022}{C_{r - 1}}} $
$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}} + \sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}} } } \right]$
$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{{2022} \over {(r - 1)}}\,.\,{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$
$ = 2023\left[ {2022\sum\limits_{r = 0}^{2023} {{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$
$ = 2023\left[ {2022\,.\,{2^{2021}} + {2^{2022}}} \right]$
$ = 2023\,.\,{2^{2021}}\left[ {2022 + 2} \right]$
$ = 2023\,.\,{2^{2021}}\,.\,2024$
$ = 2023\,.\,{{{2^{2022}}} \over {{2}}}\,.\,2024$
$ = 2023\,.\,{2^{2022}}\,.\,{1012}$
$\therefore$ $\alpha = {1012}$
$ \text { If } \sum\limits_{k=1}^{10} K^{2}\left(10_{C_{K}}\right)^{2}=22000 L \text {, then } L \text { is equal to }$ ________.
Explanation:
Given,
$\sum\limits_{k = 1}^{10} {{k^2}{{\left( {{}^{10}{C_k}} \right)}^2} = 2200\,L} $
$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{}^{10}{C_k}} \right)}^2} = 22000\,L} $
$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{{10} \over k}\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $
$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {10\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $
$ \Rightarrow 100\,.\,\sum\limits_{k = 1}^{10} {{{\left( {{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $
$ \Rightarrow 100\left( {{{\left( {{}^9{C_0}} \right)}^2} + {{\left( {{}^9{C_1}} \right)}^2}\, + \,....\, + \,{{\left( {{}^9{C_9}} \right)}^2}} \right) = 22000\,L$
$ \Rightarrow 100\left( {{}^{18}{C_9}} \right) = 22000\,L$
[Note : ${\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2}\, + \,....\, + \,{\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$]
$ \Rightarrow 100 \times {{18!} \over {9!\,9!}} = 22000\,L$
$ \Rightarrow L = 221$
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$, in the increasing powers of $\frac{1}{\sqrt[4]{3}}$ be $\sqrt[4]{6}: 1$. If the sixth term from the beginning is $\frac{\alpha}{\sqrt[4]{3}}$, then $\alpha$ is equal to _________.
Explanation:
Fifth term from beginning $ = {}^n{C_4}{\left( {{2^{{1 \over 4}}}} \right)^{n - 4}}{\left( {{3^{{{ - 1} \over 4}}}} \right)^4}$
Fifth term from end $ = {(n - 5 + 1)^{th}}$ term from begin $ = {}^n{C_{n - 4}}{\left( {{2^{{1 \over 4}}}} \right)^3}{\left( {{3^{{{ - 1} \over 4}}}} \right)^{n - 4}}$
Given ${{{}^n{C_4}{2^{{{n - 4} \over 4}}}\,.\,{3^{ - 1}}} \over {{}^n{C_{n - 3}}{2^{{4 \over 4}}}\,.\,{3^{ - \left( {{{n - 4} \over 4}} \right)}}}} = {6^{{1 \over 4}}}$
$ \Rightarrow {6^{{{n - 8} \over 4}}} = {6^{{1 \over 4}}}$
$ \Rightarrow {{n - 8} \over 4} = {1 \over 4} \Rightarrow n = 9$
${T_6} = {T_{5 + 1}} = {}^9{C_5}{\left( {{2^{{1 \over 4}}}} \right)^4}{\left( {{3^{{{ - 1} \over 4}}}} \right)^5}$
$ = {{{}^9{C_5}\,.\,2} \over {{3^{{1 \over 4}}}\,.\,3}} = {{84} \over {{3^{{1 \over 4}}}}} = {\alpha \over {{3^{{1 \over 4}}}}}$
$ \Rightarrow \alpha = 84.$
Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an A.P. If d is the common difference of this A.P. , then $50-\frac{2 d}{\beta^{2}}$ is equal to __________.
Explanation:
Coefficients of middle terms of given expansions are ${}^4{C_2}{1 \over 6}{\beta ^2},\,{}^2{C_1}( - 3\beta ),\,{}^6{C_3}{\left( {{{ - \beta } \over 2}} \right)^3}$ form an A.P.
$\therefore$ $2.2( - 3\beta ) = {\beta ^2} - {{5{\beta ^3}} \over 2}$
$ \Rightarrow - 24 = 2\beta - 5{\beta ^2}$
$ \Rightarrow 5{\beta ^2} - 2\beta - 24 = 0$
$ \Rightarrow 5{\beta ^2} - 12\beta + 10\beta - 24 = 0$
$ \Rightarrow \beta (5\beta - 12) + 2(5\beta - 12) = 0$
$\beta = {{12} \over 5}$
$d = - 6\beta - {\beta ^2}$
$\therefore$ $50 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57$
If $1 + (2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}})$ is equal to $2^{\mathrm{n}} \cdot \mathrm{m}$, where $\mathrm{m}$ is odd, then $\mathrm{n}+\mathrm{m}$ is equal to __________.
Explanation:
$l = 1 + (1 + {}^{49}{C_0} + {}^{49}{C_1}\, + \,....\, + \,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}})$
As ${}^{49}{C_0} + {}^{49}{C_1}\, + \,.....\, + \,{}^{49}{C_{49}} = {2^{49}}$
and ${}^{50}{C_0} + {}^{50}{C_2}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}}$
$ \Rightarrow {}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}} - 1$
$\therefore$ $l = 1 + ({2^{49}} + 1)({2^{49}} - 1)$
$ = {2^{98}}$
$\therefore$ $m = 1$ and $n = 98$
$m + n = 99$
Let for the $9^{\text {th }}$ term in the binomial expansion of $(3+6 x)^{\mathrm{n}}$, in the increasing powers of $6 x$, to be the greatest for $x=\frac{3}{2}$, the least value of $\mathrm{n}$ is $\mathrm{n}_{0}$. If $\mathrm{k}$ is the ratio of the coefficient of $x^{6}$ to the coefficient of $x^{3}$, then $\mathrm{k}+\mathrm{n}_{0}$ is equal to :
Explanation:
${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$
If T9 is numerically greatest term
$\therefore$ ${T_8} \le {T_9} \le {T_{10}}$
${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$
$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {(n - 8)!8!}}3\,.\,(6x) \ge {{n!} \over {(n - 9)!9!}}{(6x)^2}$
$ \Rightarrow \underbrace {{9 \over {(n - 7)(n - 8)}}}_{} \le \underbrace {{{18\left( {{3 \over 2}} \right)} \over {(n - 8)8}} \ge {{36} \over {9.8}}{9 \over 4}}_{}$
$72 \le 27(n - 7)$ and $27 \ge 9(n - 8)$
${{29} \over 3} \le n$and $n \le 11$
$\therefore$ ${n_0} = 10$
For ${(3 + 6x)^{10}}$
${T_{r + 1}} = {}^{10}{C_r}$
${3^{10 - r}}{(6x)^r}$
For coeff. of x6
$r = 6 \Rightarrow {}^{10}{C_6}{3^4}{.6^6}$
For coeff. of x3
$r = 3 \Rightarrow {}^{10}{C_3}{3^7}{.6^3}$
$\therefore$ $k = {{{}^{10}{C_6}} \over {{}^{10}{C_3}}}.{{{3^4}{{.6}^6}} \over {{3^7}{{.6}^3}}} = {{10!7!3!} \over {6!4!10!}}.8$
$ \Rightarrow k = 14$
$\therefore$ $k + {n_0} = 24$
If the coefficients of $x$ and $x^{2}$ in the expansion of $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}, \mathrm{p}, \mathrm{q} \leq 15$, are $-3$ and $-5$ respectively, then the coefficient of $x^{3}$ is equal to _____________.
Explanation:
Coefficient of x in ${(1 + x)^p}{(1 - x)^q}$
$ - {}^p{C_0}\,{}^q{C_1} + {}^p{C_1}\,{}^q{C_0} = - 3 \Rightarrow p - q = - 3$
Coefficient of x2 in ${(1 + x)^p}{(1 - x)^q}$
${}^p{C_0}\,{}^q{C_2} - {}^p{C_1}\,{}^q{C_1} + {}^p{C_2}\,{}^q{C_0} = - 5$
${{q(q - 1)} \over 2} - pq + {{p(q - 1)} \over 2} = - 5$
${{{q^2} - q} \over 2} - (q - 3)q + {{(q - 3)(q - 4)} \over 2} = - 5$
$ \Rightarrow q = 11,\,p = 8$
Coefficient of x3 in ${(1 + x)^8}{(1 - x)^{11}}$
$ = - {}^{11}{C_3} + {}^8{C_1}\,{}^{11}{C_2} - {}^8{C_2}\,{}^{11}{C_1} + {}^8{C_3} = 23$
If the maximum value of the term independent of $t$ in the expansion of $\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0$, is $\mathrm{K}$, then $8 \mathrm{~K}$ is equal to ____________.
Explanation:
General term of ${\left( {{t^2}{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{15}}$ is
${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {{t^2}{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( {{{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^r}$
$ = {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}\,.\,{t^{ - r}}$
$ = {}^{15}{C_r}\,.\,{t^{30 - 3r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}$
Term will be independent of $\mathrm{t}$ when $30 - 3r = 0 \Rightarrow r = 10$
$\therefore$ ${T_{10 + 1}} = {T_{11}}$ will be independent of $\mathrm{t}$
$\therefore$ ${T_{11}} = {}^{15}{C_{10}}\,.\,{x^{{{15 - 10} \over 5}}}\,.\,{\left( {1 - x} \right)^{{{10} \over {10}}}}$
$ = {}^{15}{C_{10}}\,.\,{x^1}\,.\,{\left( {1 - x} \right)^1}$
$\mathrm{T_{11}}$ will be maximum when $x(1 - x)$ is maximum.
Let $f(x) = x(1 - x) = x - {x^2}$
$f(x)$ is maximum or minimum when $f'(x) = 0$
$\therefore$ $f'(x) = 1 - 2x$
For maximum/minimum $f'(x) = 0$
$\therefore$ $1 - 2x = 0$
$ \Rightarrow x = {1 \over 2}$
Now, $f''(x) = - 2 < 0$
$\therefore$ At $ x = {1 \over 2}$, $f(x)$ maximum
$\therefore$ Maximum value of $\mathrm{T_{11}}$ is
$ = {}^{15}{C_{10}}\,.\,{1 \over 2}\left( {1 - {1 \over 2}} \right)$
$ = {}^{15}{C_{10}}\,.\,{1 \over 4}$
Given $K = {}^{15}{C_{10}}\,.\,{1 \over 4}$
Now, $8K = 2\left( {{}^{15}{C_{10}}} \right)$
$ = 6006$
Let the coefficients of x$-$1 and x$-$3 in the expansion of ${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}},x > 0$, be m and n respectively. If r is a positive integer such that $m{n^2} = {}^{15}{C_r}\,.\,{2^r}$, then the value of r is equal to __________.
Explanation:
Given, Binomial expansion
${\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}}$
$\therefore$ General Term
${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {2{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( { - {1 \over {{x^{{1 \over 5}}}}}} \right)^r}$
$ = {}^{15}{C_r}\,.\,{2^{15 - r}}\,.\,{x^{{1 \over 5}(15 - r - r)}}\,.\,{( - 1)^r}$
For ${x^{ - 1}}$ term;
${1 \over 5}(15 - 2r) = - 1$
$ \Rightarrow 15 - 2r = - 5$
$ \Rightarrow 2r = 20$
$ \Rightarrow r = 10$
m is the coefficient of ${x^{ - 1}}$ term,
$\therefore$ $m = {}^{15}{C_{10}}\,.\,{2^{15 - 10}}\,.\,{( - 1)^{10}}$
$ = {}^{15}{C_{10}}\,.\,{2^5}$
For ${x^{ - 3}}$ term;
${1 \over 5}(15 - 2r) = - 3$
$ \Rightarrow 15 - 2r = - 15$
$ \Rightarrow 2r = 30$
$ \Rightarrow r = 15$
n is the coefficient of ${x^{ - 3}}$ term,
$\therefore$ $n = {}^{15}{C_{15}}\,.\,{2^{15 - 15}}\,.\,{( - 1)^{15}}$
$ = 1\,.\,1\,.\, - 1$
$ = - 1$
Given,
$m{n^2} = {}^{15}{C_r}\,.\,{2^r}$
$ \Rightarrow {}^{15}{C_{10}}\,.\,{2^5}\,.\,{(1)^2} = {}^{15}{C_r}\,.\,{2^r}$ [putting value of m and n]
$ \Rightarrow {}^{15}{C_{15 - 10}}\,.\,{2^5} = {}^{15}{C_r}\,.\,{2^r}$
$ \Rightarrow {}^{15}{C_5}\,.\,{2^5} = {}^{15}{C_r}.\,{2^r}$
Comparing both side, we get
$r = 5$.
The number of positive integers k such that the constant term in the binomial expansion of ${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$, x $\ne$ 0 is 28 . l, where l is an odd integer, is ______________.
Explanation:
Given Binomial expression is
${\left( {2{x^3} + {3 \over {{x^k}}}} \right)^{12}}$
General term,
${T_{r + 1}} = {}^{12}{C_r}{(2{x^3})^r}\,.\,{\left( {{3 \over {{x^k}}}} \right)^{12 - r}}$
$ = \left( {{}^{12}{C_r}\,.\,{2^r}\,.\,{3^{12 - r}}} \right)\,.\,{x^{3r - 12k + kr}}$
For constant term,
$3r - 12k + kr = 0$
$ \Rightarrow k(12 - r) = 3r$
$ \Rightarrow k = {{3r} \over {12 - r}}$
For r = 1, $k = {3 \over {11}}$ (not integer)
For r = 2, $k = {6 \over {10}}$ (not integer)
For r = 3, $k = {9 \over {9}}=1$ (integer)
For r = 6, $k = {18 \over {6}}=3$ (integer)
For r = 8, $k = {24 \over {4}}=6$ (integer)
For r = 9, $k = {27 \over {3}}=9$ (integer)
For r = 10, $k = {30 \over {2}}=15$ (integer)
For r = 11, $k = {33 \over {1}}=33$ (integer)
So, for r = 3, 6, 8, 9, 10 and 11 k is positive integer.
When k = 1 then r = 3 and constant term is
$ = {}^{12}{C_3}\,.\,{2^3}\,.\,{3^9}$
$ = {{12\,.\,11\,.\,10} \over {3\,.\,2\,.\,1}}\,.\,{2^3}\,.\,{3^9}$
$ = 2\,.\,11\,.\,2\,.\,5\,.\,{2^3}\,.\,{3^9}$
$ = 11\,.\,5\,.\,{2^5}\,.\,{3^9}$
$ = {2^5}\,.\,(55\,.\,{3^9})$
$ = {2^5}(l)$
$ \ne {2^8}\,.\,l$
When x = 3 then r = 6 and constant term
$ = {}^{12}{C_6}\,.\,{2^6}\,.\,{3^6}$
$ = {{12\,.\,11\,.\,10\,.\,9\,.\,8\,.\,7} \over {6\,.\,5\,.\,4\,.\,3\,.\,2\,.\,1}}\,.\,{2^6}\,.\,{3^6}$
$ = {2^8}\,.\,231\,.\,{3^6}$
$ = {2^8}(l)$
When k = 6 then r = 8 and constant term
$ = {}^{12}{C_8}\,.\,{2^8}\,.\,{3^4}$
$ = {{12\,.\,11\,.\,10\,.\,9} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{2^8}\,.\,{3^4}$
$ = {2^8}\,.\,55\,.\,{3^6}$
$ = {2^8}(l)$
When x = 9 then r = 9 and constant term
$ = {}^{12}{C_9}\,.\,{2^9}\,.\,{3^3}$
$ = {{12\,.\,11\,.\,10} \over {3\,.\,2\,.\,1}}\,.\,{2^9}\,.\,{3^3}$
$ = {2^{11}}\,.\,55\,.\,{3^3}$
Here power of 2 is 11 which is greater than 8. So, k = 9 is not possible.
Similarly for k = 15 and k = 33, ${2^8}\,.\,l$ form is not possible.
$\therefore$ k = 3 and k = 6 is accepted.
$\therefore$ For 2 positive integer value of k, ${2^8}\,.\,l$ form of constant term possible.
If the sum of the coefficients of all the positive powers of x, in the Binomial expansion of ${\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$ is 939, then the sum of all the possible integral values of n is _________.
Explanation:
Given, Binomial expression is
$ = {\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$
$\therefore$ General term
${T_{r + 1}} = {}^7{C_r}\,.\,{({x^n})^{7 - r}}\,.\,{\left( {{2 \over {{x^5}}}} \right)^r}$
$ = {}^7{C_r}\,.\,{x^{7n - nr - 5r}}\,.\,{2^r}$
For positive power of x,
$7n - nr - 5r > 0$
$ \Rightarrow 7n > r(n + 5)$
$ \Rightarrow r < {{7n} \over {n + 5}}$
As r represent term of binomial expression so r is always integer.
Given that sum of coefficient is 939.
When $r = 0$,
sum of coefficient $ = {}^7{C_0}\,.\,{2^0} = 1$
when $r = 1$,
sum of coefficient $ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} = 1 + 14 = 15$
when $r = 2$,
sum of coefficient
$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2}$
$ = 1 + 14 + 84$
$ = 99$
when $r = 3$,
sum of coefficient
$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2} + {}^7{C_3}\,.\,{2^3}$
$ = 1 + 14 + 84 + 280$
$ = 379$
when $r = 4$,
sum of coefficient
$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2} + {}^7{C_3}\,.\,{2^3} + {}^7{C_4}\,.\,{2^4}$
$ = 1 + 14 + 84 + 280 + 560$
$ = 939$
$\therefore$ For r = 4 sum of coefficient = 939
To get value of r = 4, value of ${{7n} \over {n + 5}}$ should be between 4 and 5.
$\therefore$ $4 < {{7n} \over {n + 5}} < 5$
$ \Rightarrow 4n + 20 < 7n < 5n + 25$
$\therefore$ $4n + 20 < 7n$
$ \Rightarrow 3n > 20$
$ \Rightarrow n > {{20} \over 3}$
$ \Rightarrow n > 6.66$
and
$7n < 5n + 25$
$ \Rightarrow 2n < 25$
$ \Rightarrow n < 12.5$
$\therefore$ $6.66 < n < 12.5$
$\therefore$ Possible integer values of $n = 7,8,9,10,11,12$
$\therefore$ Sum of values of $n = 7 + 8 + 9 + 10 + 11 + 12$
$ = 57$
If the coefficient of x10 in the binomial expansion of ${\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt 5 } \over {{x^{{1 \over 3}}}}}} \right)^{60}}$ is ${5^k}\,.\,l$, where l, k $\in$ N and l is co-prime to 5, then k is equal to _____________.
Explanation:
Given Binomial Expansion
$ = {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}}$
$\therefore$ General term
${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/3}}}}} \right)^r}$
$ = {}^{60}{C_r}\,.\,{5^{\left( {{r \over 4} - 15 + {r \over 2}} \right)}}\,.\,{x^{\left( {30 - {r \over 2} - {r \over 3}} \right)}}$
$ = {}^{60}{C_r}\,.\,{5^{\left( {{{3r - 60} \over 4}} \right)}}\,.\,{x^{\left( {{{180 - 5r} \over 6}} \right)}}$
For x10 term,
${{180 - 5r} \over 6} = 10$
$ \Rightarrow 5r = 120$
$ \Rightarrow r = 24$
$\therefore$ Coefficient of ${x^{10}} = {}^{60}{C_{24}}\,.\,{5^{\left( {{{3 \times 24 - 60} \over 4}} \right)}}$
$ = {}^{60}{C_{24}}\,.\,{5^3}$
$ = {{60!} \over {24!\,\,36!}}\,.\,{5^3}$
It is given that,
${{60!} \over {24!\,\,36!}}\,.\,{5^3} = {5^k}\,.\,l$ ...... (1)
Also given that, l is coprime to 5 means l can't be multiple of 5. So we have to find all the factors of 5 in 60!, 24! and 36!
[Note : Formula for exponent or degree of prime number in n!.
Exponent of p in $n! = \left\lceil {{n \over p}} \right\rceil + \left\lceil {{n \over {{p^2}}}} \right\rceil + \left\lceil {{n \over {{p^3}}}} \right\rceil + $ ..... until 0 comes
here p is a prime number. ]
$\therefore$ Exponent of 5 in 60!
$= \left\lceil {{{60} \over 5}} \right\rceil + \left\lceil {{{60} \over {{5^2}}}} \right\rceil + \left\lceil {{{60} \over {{5^3}}}} \right\rceil + $ .....
$ = 12 + 2 + 0 + $ .....
$ = 14$
Exponent of 5 in 24!
$ = \left\lceil {{{24} \over 5}} \right\rceil + \left\lceil {{{24} \over {{5^2}}}} \right\rceil + \left\lceil {{{24} \over {{5^3}}}} \right\rceil + $ ......
$ = 4 + 0 + 0$ ......
$ = 4$
Exponent of 5 in 36!
$ = \left\lceil {{{36} \over 5}} \right\rceil + \left\lceil {{{36} \over {{5^2}}}} \right\rceil + \left\lceil {{{36} \over {{5^3}}}} \right\rceil + $ .......
$ = 7 + 1 + 0$ ......
$ = 8$
$\therefore$ From equation (1), exponent of 5 overall
${{{5^{14}}} \over {{5^4}\,.\,{5^8}}}\,.\,{5^3} = {5^k}$
$ \Rightarrow {5^5} = {5^k}$
$ \Rightarrow k = 5$
Explanation:
Here property used is
${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$
Given, ${}^{40}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + \,\,....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}{}^{60}{C_{20}}$
As ${}^{40}{C_0} = {}^{41}{C_0} = 1$
So, we replace ${}^{40}{C_0}$ with ${}^{41}{C_0}$.
$ \Rightarrow {}^{41}{C_0} + {}^{41}{C_1} + {}^{42}{C_2} + \,\,.....\,\,\, + \,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \Rightarrow {}^{42}{C_1} + {}^{42}{C_2} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \Rightarrow {}^{43}{C_2} + {}^{43}{C_3} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \Rightarrow {}^{44}{C_3} + {}^{44}{C_4} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \Rightarrow {}^{45}{C_4} + {}^{45}{C_5} + \,\,.....\,\, + \,\,{}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \vdots $
$ \Rightarrow {}^{60}{C_{19}} + {}^{60}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \Rightarrow {}^{61}{C_{20}} = {m \over n}\,.\,{}^{60}{C_{20}}$
$ \Rightarrow {{61!} \over {20!\,41!}} = {m \over n}\,.\,{{60!} \over {20!\,40!}}$
$ \Rightarrow {{61} \over {41}} = {m \over n}$
$\therefore$ m = 61 and n = 41
$\therefore$ m + n = 61 + 41 = 102
If the sum of the co-efficient of all the positive even powers of x in the binomial expansion of ${\left( {2{x^3} + {3 \over x}} \right)^{10}}$ is ${5^{10}} - \beta \,.\,{3^9}$, then $\beta$ is equal to ____________.
Explanation:
Given, Binomial Expansion
${\left( {2{x^3} + {3 \over x}} \right)^{10}}$
General term
${T_{r + 1}} = {}^{10}{C_r}\,.\,{(2{x^3})^{10 - r}}\,.\,{\left( {{3 \over x}} \right)^r}$
$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 3r}}\,.\,{x^{ - r}}$
$ = {}^{10}{C_r}\,.\,{2^{10 - r}}\,.\,{3^r}\,.\,{x^{30 - 4r}}$
For positive even power of x, 30 $-$ 4r should be even and positive.
For r = 0, 30 $-$ 4 $\times$ 0 = 30 (even and positive)
For r = 1, 30 $-$ 4 $\times$ 1 = 26 (even and positive)
For r = 2, 30 $-$ 4 $\times$ 2 = 22 (even and positive)
For r = 3, 30 $-$ 4 $\times$ 3 = 18 (even and positive)
For r = 4, 30 $-$ 4 $\times$ 4 = 14 (even and positive)
For r = 5, 30 $-$ 4 $\times$ 5 = 10 (even and positive)
For r = 6, 30 $-$ 4 $\times$ 6 = 6 (even and positive)
For r = 7, 30 $-$ 4 $\times$ 7 = 2 (even and positive)
For r = 8, 30 $-$ 4 $\times$ 8 = $-$2 (even but not positive)
So, for r = 1, 2, 3, 4, 5, 6 and 7 we can get positive even power of x.
$\therefore$ Sum of coefficient for positive even power of x
$ = {}^{10}{C_0}\,.\,{2^{10}}\,.\,{3^0} + {}^{10}{C_1}\,.\,{2^9}\,.\,{3^1} + {}^{10}{C_2}\,.\,{2^8}\,.\,{3^2} + {}^{10}{C_3}\,.\,{2^7}\,.\,{3^3} + {}^{10}{C_4}\,.\,{2^6}\,.\,{3^4} + {}^{10}{C_5}\,.\,{2^5}\,.\,{3^5} + {}^{10}{C_6}\,.\,{2^4}\,.\,{3^6} + {}^{10}{C_7}\,.\,{2^3}\,.\,{3^7}$
$ = {}^{10}{C_{10}}\,.\,{2^{10}}\,.\,{3^0} + {}^{10}{C_1}\,.\,{2^9}\,.\,{3^1}\,\, + \,\,.....\,\, + \,\,{}^{10}{C_{10}}\,.\,{2^0}\,.\,{3^{10}} - \left[ {{}^{10}{C_8}\,.\,{2^2}\,.\,{3^8} + {}^{10}{C_9}\,.\,2\,.\,{3^9} + {}^{10}{C_{10}}\,.\,{2^0}\,.\,{3^{10}}} \right]$
$ = {(2 + 3)^{10}} - \left[ {45\,.\,4\,.\,{3^8} + 10\,.\,2\,.\,{3^9} + 1\,.\,1\,.\,{3^{10}}} \right]$
$ = {5^{10}} - \left[ {60 \times {3^9} + 20\,.\,{3^9} + 3\,.\,{3^9}} \right]$
$ = {5^{10}} - \left( {60 + 20 + 3} \right){3^9}$
$ = {5^{10}} - 83\,.\,{3^9}$
$\therefore$ $\beta = 83$
Let Cr denote the binomial coefficient of xr in the expansion of ${(1 + x)^{10}}$. If for $\alpha$, $\beta$ $\in$ R, ${C_1} + 3.2{C_2} + 5.3{C_3} + $ ....... upto 10 terms $ = {{\alpha \times {2^{11}}} \over {{2^\beta } - 1}}\left( {{C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + \,\,.....\,\,upto\,10\,terms} \right)$ then the value of $\alpha$ + $\beta$ is equal to ___________.
Explanation:
Given,
${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $ ...... upto 10 terms
$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3}$ + ..... upto 10 terms)
Now,
L.H.S. :-
${C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $ ...... upto 10 terms
$ = 1\,.\,1{C_1} + 3\,.\,2{C_2} + 5\,.\,3{C_3} + $ ..... upto 10 terms
$ = \sum\limits_{r = 1}^{10} {r\,.\,(2r - 1){}^{10}{C_r}} $
$ = \sum\limits_{r = 1}^{10} {(2{r^2} - r)\,.\,{}^{10}{C_r}} $
$ = 2\,.\,\sum\limits_{r = 1}^{10} {{r^2}\,.\,{}^{10}{C_r} - \sum\limits_{r = 1}^{10} {r\,.\,{}^n{C_r}} } $
[We know, $\sum\limits_{r = 1}^n {r\,.\,{}^n{C_r} = n\,.\,{2^{n - 1}}} $
and $\sum\limits_{r = 1}^n {{r^2}\,.\,{}^n{C_r} = \sum\limits_{r = 1}^n {(r\,.\,{}^n{C_r})\,.\,r} } $
$ = \sum\limits_{r = 1}^n {\left( {r\,.\,{n \over r}\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r} $
$ = \sum\limits_{r = 1}^n {\left( {n\,.\,{}^{n - 1}{C_{r - 1}}} \right)\,.\,r} $
$ = n\sum\limits_{r = 1}^n {(r - 1 + 1){}^{n - 1}{C_{r - 1}}} $
$ = n\,.\,\sum\limits_{r = 1}^n {(r - 1)\,.\,{}^{n - 1}{C_{r - 1}} + n\,.\,\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} } $
$ = n\,.\,(n - 1)\,.\,{2^{n - 2}} + n\,.\,{2^{n - 1}}$]
$ = 2\left( {n(n - 1){2^{n - 2}} + n\,.\,{2^{n - 1}}} \right) - n\,.\,{2^{n - 1}}$
Put $n = 10$
$ = 2\left( {10\,.\,9\,.\,{2^8} + 10\,.\,{2^9}} \right) - 10\,.\,{2^9}$
$ = 45\,.\,{2^{10}} + 10\,.\,{2^{10}} - 5\,.\,{2^{10}}$
$ = {2^{10}}(45 + 10 - 5)$
$ = {2^{10}}\,.\,(50)$
$ = 25\,.\,{2^{11}}$
R.H.S. :-
${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${C_0} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + $ ..... upto 10 terms)
${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}$ (${{{C_0}} \over 1} + {{{C_1}} \over 2} + {{{C_2}} \over 3} + $ ..... upto 10 terms)
${{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^n{C_r}} \over {r + 1}}} } \right)$
$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {\sum\limits_{r = 0}^n {{{{}^{n + 1}{C_{r + 1}}} \over {n + 1}}} } \right)$
$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_1} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}}} \over {n + 1}}} \right)$
$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{}^{n + 1}{C_0} + {}^{n + 1}{C_2} + \,\,....\,\, + \,\,{}^{n + 1}{C_{n + 1}} - {}^{n + 1}{C_0}} \over {n + 1}}} \right)$
$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{n + 1}} - 1} \over {n + 1}}} \right)$
Putting value of $n = 10$, we get
$ = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$
Using L.H.S. = R.H.S.
$ \Rightarrow 25\,.\,{2^{11}} = {{\alpha \,.\,{2^{11}}} \over {{2^\beta } - 1}}\left( {{{{2^{11}} - 1} \over {11}}} \right)$
$ \Rightarrow 25\,.\,{2^{11}} = {2^{11}}\left( {{\alpha \over {11}}} \right)\left( {{{{2^{11}} - 1} \over {{2^\beta } - 1}}} \right)$
By comparing both sides,
${\alpha \over {11}} = 25 \Rightarrow \alpha = 275$
and ${{{2^{11}} - 1} \over {{2^\beta } - 1}} = 1$
$ \Rightarrow {2^{11}} = {2^\beta }$
$ \Rightarrow \beta = 11$
$\therefore$ $\alpha + \beta = 275 + 11 = 286$
The remainder on dividing 1 + 3 + 32 + 33 + ..... + 32021 by 50 is _________.
Explanation:
Given,
$1 + 3 + {3^2} + {3^3} + \,\,.....\,\, + \,\,{3^{2021}}$
$ = {3^0} + {3^1} + {3^2} + {3^3} + \,\,....\,\, + \,\,{3^{2021}}$
This is a G.P with common ratio = 3
$\therefore$ Sum $ = {{1({3^{2022}} - 1)} \over {3 - 1}}$
$ = {{{3^{2022}} - 1} \over 2}$
$ = {{{{({3^2})}^{2011}} - 1} \over 2}$
$ = {{{{(10 - 1)}^{1011}} - 1} \over 2}$
$ = {{\left[ {{}^{1011}{C_0}\,.\,{{10}^{1011}} - {}^{1011}{C_1}\,.\,{{10}^{1010}} + \,\,.....\,\, - \,\,{}^{1011}{C_{1009}}\,.\,{{(10)}^2} + {}^{1011}{C_{1010}}\,.\,10 - {}^{1011}{C_{1011}}} \right] - 1} \over 2}$
$ = {{{{10}^2}\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1009}} - {}^{1011}{C_1}\,.\,(1008) + \,\,.....\,\,{}^{1011}{C_{1009}}} \right] + 10110 - 1 - 1} \over 2}$
$ = {{100k + 10110 - 2} \over 2}$
$ = {{100k + 10108} \over 2}$
$ = 50k + 5054$
$ = 50k + 50 \times 101 + 4$
$ = 50[k + 101] + 4$
$ = 50k' + 4$
$\therefore$ By dividing 50 we get remainder as 4.
Explanation:
210 = 1024 $\times$ 2
$\Rightarrow$ 2n = 212
211 = 2048
n = 12
212 = 4096
${}^{12}{C_6}={{12 \times 11 \times 10 \times 9 \times 8 \times 7} \over {6 \times 5 \times 4 \times 3 \times 2 \times 1}}$
$ = 11 \times 3 \times 4 \times 7$
$ = 924$
Explanation:
${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$
$\alpha + \beta + \gamma = 10$ ..... (1)
$\alpha + \gamma = 7$ .... (2)
$\beta + \gamma = 8$ ..... (3)
$(2) + (3) - (1) \Rightarrow \gamma = 5$
$\alpha = 2$
$\beta = 3$
so coefficients = ${{10!} \over {2!3!5!}}{2^3}{.2^{10}}$
$ = {{10 \times 9 \times 8 \times 7 \times 6 \times 5} \over {2 \times 3 \times 2 \times 5!}} \times {2^{13}}$
$ = 315 \times {2^{16}} \Rightarrow k = 315$
Explanation:
${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$
${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{12 - 3r}}$
Term independent of x $\Rightarrow$ 12 $-$ 3r = 0 $\Rightarrow$ r = 4
${T_5} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^8}{\left( {12} \right)^4} = {{{3^6}} \over {{4^4}}}.\,k$
$\Rightarrow$ k = 55
Explanation:
= $-$ 39 + 18 . I
= (54 $-$ 39) + 18(I $-$ 3)
= 15 + 18I1
$\Rightarrow$ Remainder = 15
If ${A_k} = \sum\limits_{i = 0}^9 {\left( {\matrix{ 9 \cr i \cr } } \right)\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right] + } \sum\limits_{i = 0}^8 {\left( {\matrix{ 8 \cr i \cr } } \right)\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right]} $ and A4 $-$ A3 = 190 p, then p is equal to :
Explanation:
${A_k} = {}^{21}{C_k} + {}^{21}{C_k} = 2.{}^{21}{C_k}$
${A_4} - {A_3} = 2\left( {{}^{21}{C_4} - {}^{21}{C_3}} \right) = 2(5985 - 1330)$
$190p = 2(5985 - 1330) \Rightarrow p = 49$
Explanation:
${T_r} = (2r + 1){}^n{C_r}$
$S = \sum {{T_r}} $
$S = \sum {(2r + 1){}^n{C_r}} = \sum {2r{}^n{C_r} + \sum {{}^n{C_r}} } $
$S = 2(n{.2^{n - 1}}) + {2^n} = {2^n}(n + 1)$
${2^n}(n + 1) = {2^{100}}.101 \Rightarrow n = 100$
$2\left[ {{{n - 1} \over 2}} \right] = 2\left[ {{{99} \over 2}} \right] = 98$
Explanation:
$\Rightarrow$ n $-$ 7 = 48 $\Rightarrow$ n = 55
Explanation:
So required ratio = ${{{}^{20}{C_{10}}} \over {^{19}{C_9}{ + ^{19}}{C_{10}}}} = {{^{20}{C_{10}}} \over {^{20}{C_{10}}}} = 1$
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$, where x $\ne$ 0, 1 is equal to ______________.
Explanation:
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
[Note:
For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is
$r = {{n\alpha - m} \over {\alpha + \beta }}$]
Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = 0
then $r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$ = ${{10} \over 3} \times {6 \over 5}$ = 4
$\therefore$ T5 is the term independent of x.
$\therefore$ T5 = ${}^{10}{C_4}$ = 210
Explanation:
General term $ = {}^{10}{C_R}{(2{x^2})^{10 - R}}{x^{ - 2R}}$
$ \Rightarrow {2^{10 - R}}{}^{10}{C_R} = 180$ ....... (1)
& (10 $-$ R)r $-$ 2R = 0
$r = {{2R} \over {10 - R}}$
$r = {{2(R - 10)} \over {10 - R}} + {{20} \over {10 - R}}$
$ \Rightarrow r = - 2 + {{20} \over {10 - R}}$ ....... (2)
R = 8 or 5 reject equation (1) not satisfied
At R = 8
$ \Rightarrow {2^{10 - R}}\times{}^{10}{C_R} = 180 \Rightarrow r = 8$
Explanation:
$ \Rightarrow {11^n} - {9^n} > {10^n}$
$ \Rightarrow {(10 + 1)^n} - {(10 - 1)^n} > {10^n}$
$ \Rightarrow 2\{ {}^n{C_1}{.10^{n - 1}} + {}^n{C_3}{10^{n - 10}} + {}^n{C_5}{10^{n - 5}} + .....\} > {10^n}$
$ \Rightarrow $ ${1 \over 5}\left[ {{}^n{C_1}{{10}^n} + {}^n{C_3}{{10}^{n - 2}} + {}^n{C_5}{{10}^{n - 4}} + .....} \right] > {10^n}$
$ \Rightarrow $ ${1 \over 5}\left[ {{}^n{C_1} + {}^n{C_3}{{10}^{ - 2}} + {}^n{C_5}{{10}^{ - 4}} + .....} \right] > 1$
Clearly the above inequality is true for n $ \ge $ 5
For n = 4, we have ${1 \over 5}\left[ {4 + {4 \over {{{10}^2}}}} \right] = {4 \over 5}\left( {{{101} \over {100}}} \right) < 1$
$\Rightarrow$ Inequality does not hold good for n = 1, 2, 3, 4
So, required number of elements ={5, 6, 7, ......., 100} = 96
Explanation:
${T_{r + 1}} = {}^{120}{C_r}{({2^{1/2}})^{120 - r}}{(5)^{r/6}}$
for rational terms r = 6$\lambda$
0 $\le$ r $\le$ 120
So total no of terms are 21.
${\left[ {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right]^{10}}$, x $\ne$ 1, is equal to ____________.
Explanation:
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
${T_{r + 1}} = {}^{10}{C_r}{\left( {{x^{{1 \over 3}}}} \right)^{(10 - r)}}{\left( {{x^{ - {1 \over 2}}}} \right)^r}$
For being independent of $x:{{10 - r} \over 3} - {r \over 2} = 0 \Rightarrow r = 4$
Term independent of $x = {}^{10}{C_4} = 210$
Explanation:
$ = 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k.{}^{10}{C_k}} $
$ = 4({2^{10}}) + 3\sum\limits_{k = 0}^{10} {k.{{10} \over k}.{}^9{C_{k - 1}}} $
= $4({2^{10}}) + 3.10({2^9})$
$ = 4({2^{10}}) + {3.5.2^{10}}$
$ = {2^{10}}(19)$
According to question,
$19({2^{10}}) = \alpha {.3^{10}} + \beta {.2^{10}}$
$ \therefore $ $\alpha = 0,\beta = 19$
$ \Rightarrow \alpha + \beta = 19$
Explanation:
$ = {}^n{C_r}{a^r}{x^{n - 3r}}$
${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$, ${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$, ${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$
Now, ${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}$
$ \Rightarrow a(n - 2) = 2$ .......... (i)
and ${{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}$
$ \Rightarrow a(n - 3) = {3 \over 2}$ ........ (ii)
by (i) and (ii) $n = 6,\,a = {1 \over 2}$
for term independent of 'x'
$n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2$
${T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4$
Explanation:
(2021)3762 = (17m $-$ 2)3762 = multiple of 17 + 23762
= 17$\lambda$ + 22 (24)940
= 17$\lambda$ + 4 (17 $-$ 1)940
= 17$\lambda$ + 4 (17$\mu$ + 1)
= 17k + 4; (k $\in$ I)
$ \therefore $ Remainder = 4
$A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}\left[ {{{\left( {{1 \over 2}} \right)}^k} + {{\left( {{3 \over 4}} \right)}^k} + {{\left( {{7 \over 8}} \right)}^k} + {{\left( {{{15} \over {16}}} \right)}^k} + {{\left( {{{31} \over {32}}} \right)}^k}} \right]} $. If
$63A = 1 - {1 \over {{2^{30}}}}$, then n is equal to _____________.
Explanation:
$ = {\left( {1 - {1 \over 2}} \right)^n} + {\left( {1 - {3 \over 4}} \right)^n} + ..... + {\left( {1 - {{31} \over {32}}} \right)^n}$
$ = {\left( {{1 \over 2}} \right)^n} + {\left( {{1 \over 2}} \right)^{2n}} + {\left( {{1 \over 2}} \right)^{3n}} + ..... + {\left( {{1 \over 2}} \right)^{5n}}$
$ = {\left( {{1 \over 2}} \right)^n}\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{5n}}} \over {1 - {{\left( {{1 \over 2}} \right)}^n}}}} \right) = {{{2^{5n}} - 1} \over {{2^{5n}}({2^n} - 1)}}$
$ \therefore $ $63A = {{63\left( {{2^{5n}} - 1} \right)} \over {{2^{5n}}\left( {{2^n} - 1} \right)}}$ = ${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$
Given, $63A = 1 - {1 \over {{2^{30}}}}$
$ \therefore $ ${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$ = $1 - {1 \over {{2^{30}}}}$
For n = 6, L.H.S = R.H.S
$ \therefore $ n = 6
(Here $\left( {\matrix{ n \cr k \cr } } \right) = {}^n{C_k}$)
Explanation:
$ = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$
$ = \sum\limits_{r = 1}^{30} {r({}^{30}{C_r})} $
$ = \sum\limits_{r = 1}^{30} {r\left( {{{30} \over r}} \right)({}^{29}{C_{r - 1}}} )$
$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}} $
$ = 30({}^{29}{C_0} + {}^{29}{C_1} + {}^{29}{C_2} + ..... + {}^{29}{C_{29}})$
$ = 30({2^{29}}) = 15{(2)^{30}} = n{(2)^m}$
$ \therefore $ n = 15, m = 30
$ \Rightarrow $ n + m = 45
Explanation:
(2020 + x)2022
= (2020 + 4k + 3)2022
= (4(505) + 4k + 3)2022
= (4P + 3)2022
= (4P + 4 $-$ 1)2022
= (4A $-$ 1)2022
2022C0(4A)0($-$1)2022 + 2022C1(4A)1($-$1)2021 + ......
= 1 + 2022(4A)(-1) + .....
= 1 + 8$\lambda$
$ \therefore $ Reminder is 1.
Explanation:
${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$
$ \therefore $ 3n = 32t = (10 $-$ 1)t
= 10p + ($-$1)t
= 10p $\pm$ 1
$ \therefore $ if n = even then 7n + 3n will not be multiply of 10
So if n is odd then only 7n + 3n will be multiply of 10
$ \therefore $ n = 11, 13, 15, ..........., 99
$ \therefore $ Ans : 45
Explanation:
Question will be BONUS.
(1 + x + x2 + x3)6 in powers of x, is ______.
Explanation:
= ((1 + x) (1 + x2))6
= (1 + x)6(1 + x2)6
= $\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^r}} $ $\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^{2r}}} $
Coefficient of x4 = 6C0 6C2 + 6C2 6C1 + 6C4 6C0 = 120
${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$ is 1540, is .............
Explanation:
${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$
$ \because $ ${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$
$ \therefore $ $r = 3\,or\,19$
$22m - mr - 2r = 1$
$m = {{2r + 1} \over {22 - 5}}$
When $r = 3$, $m = {7 \over {19}} \notin N$
When $r = 19$, $m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13$
$ \therefore $ $m = 13$
Then ${{{a_7}} \over {{a_{13}}}}$ is equal to ______.
Explanation:
The general term of ${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$ the expansion is
${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$
where n1 + n2 + ..... + nn = n
Here, in ${(2{x^2} + 3x + 4)^{10}}$ general term is
$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{(2{x^2})^{{n_1}}}{(3x)^{{n_2}}}{(4)^{{n_3}}}$
$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}.{x^{2{n_1} + {n_2}}}$
$ \therefore $ Coefficient of $ {x^{2{n_1} + {n_2}}}$ is
${{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}$
where ${n_1} + {n_2} + {n_3} = 10$
For, Coefficient of x7 :
2n1 + n2 = 7
Possible values of n1, n2 and n3 are
| ${n_1}$ | ${n_2}$ | ${n_3}$ |
|---|---|---|
| 3 | 1 | 6 |
| 2 | 3 | 5 |
| 1 | 5 | 4 |
| 0 | 7 | 3 |
$ \therefore $ Coefficient of x7
$ = {{10!} \over {3!1!6!}}{(2)^3}{(3)^1}{(4)^6} + {{10!} \over {3!1!6!}}{(2)^2}{(3)^3}{(4)^5} + {{10!} \over {1!5!4!}}{(2)^1}{(3)^5}{(4)^4} + {{10!} \over {0!7!3!}}{(2)^0}{(3)^7}{(4)^3}$
Coefficient of x13 = a13
Here 2n1 + n2 = 13
possible values of n1, n2 and n3 are
| ${n_1}$ | ${n_2}$ | ${n_3}$ |
|---|---|---|
| 6 | 1 | 3 |
| 5 | 3 | 2 |
| 4 | 5 | 1 |
| 3 | 7 | 0 |
$ \therefore $ Coefficient of x13
$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$
$ \therefore $ ${{{a_7}} \over {{a_{13}}}} = 8$
${\left( {1 + {1 \over x}} \right)^n}$ is expanded
in increasing powers of x. If three consecutive
coefficients in this expansion are in the ratio,
2 : 5 : 12, then n is equal to________.
Explanation:
${}^n{C_{r - 1}},{}^n{C_r},{}^n{C_{r + 1}}$
${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:5:12$
Now, ${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over 5}$
$ \Rightarrow 7r = 2n + 2$ ...(i)
${{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {5 \over {12}}$
$ \Rightarrow 7r = 5n - 12$ ...(ii)
On solving (i) and (ii) we get n = 118
C0 + 5.C1 + 9.C2 + .... + (101).C25 = 225.k, then k is equal to _____.
Explanation:
S = (101).25C25 + (97).25C24 + .......... + (1).25C0
_________________________________________
2S = 102{25C0 + 25C1 + ......+ 25C25}
$ \Rightarrow $ S = 51 $ \times $ 225 = k.225
$ \Rightarrow $ k = 51
Explanation:
= 10C0(1 + x)10 + 10C1(1 + x)9.x2 + 10C2(1 + x)8.x4+ .....
Coefficient of x4
= 10C0.10C4 + 10C1.9C2 + 10C2.8C0
= 210 + 360 + 45
= 615
(1 + x + x2 + ....+ x2n)(1 - x + x2 - x3 + ...... + x2n) is 61, then n is equal to _______.
Explanation:
= a0 + a1x + a2x2 + …..
put x = 1
(2n + 1)$ \times $1 = a0 + a1 + a2 + …… (1)
put x = –1
1$ \times $(2n + 1) = a0 – a1 + a2+ …….. (2)
Adding (1) and (2)
4n + 2 = 2(a0 + a2 + ….. )
$ \Rightarrow $ 4n + 2 = 2 $ \times $ 61
$ \Rightarrow $ n = 30