Binomial Theorem
Explanation:
$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\ & ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r} \end{aligned} $
For constant term, power of x is zero.
So, $\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45$
Now constant term $=-84$
and ${ }^{9} C_{r} \cdot 2^{3 r-9}(-1)^{r}=-84$
So, $r=3$ and $l=5$
Now for $x^{-15}, \frac{45-5 r}{2}-5 r=-15$
$ \Rightarrow $ $ 45-15 r=-30 $
$ \Rightarrow $ $ r=5 $
$\therefore $ Coefficient $=-{ }^{9} C_{5} 2^{6}=-63.2^{7}$
$\therefore \alpha=7, \beta=-63$
and $|\alpha l-\beta|=|7 \times 5+63|=98$
The remainder on dividing $5^{99}$ by 11 is ____________.
Explanation:
$=625\left[5^{5}\right]^{19}$
$=625[3125]^{19}$
$=625[3124+1]^{19}$
$=625[11 \mathrm{k} \times 19+1]$
$=625 \times 11 \mathrm{k} \times 19+625$
$=11 \mathrm{k}_{1}+616+9$
$=11\left(\mathrm{k}_{2}\right)+9$
Remainder $=9$
Let $\alpha>0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$ has a term $\beta x^{-\alpha}, \beta \in \mathbb{N}$. Then $\alpha$ is equal to ___________.
Explanation:
$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
$\frac{60-11 \mathrm{r}}{3}<0 $
$\Rightarrow 11 \mathrm{r}>60 $
$\Rightarrow \mathrm{r}>\frac{60}{11} $
$\Rightarrow \mathrm{r}=6$
$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
$\therefore \alpha=2$
Explanation:
Given ${x^{{1 \over {50}}}} = 12 \Rightarrow x = {12^{50}}$
${y^{{1 \over {50}}}} = 18 \Rightarrow y = {18^{50}}$
$12\equiv13$ (Mod 25)
$12^2\equiv19$ (Mod 25)
$12^3\equiv-3$ (Mod 25)
$12^9\equiv-2$ (Mod 25)
$12^{10}\equiv-1$ (Mod 25)
$12^{50}\equiv-1$ (Mod 25) ..... (i)
Now
$18\equiv7$ (Mod 25)
$18^2\equiv-1$ (Mod 25)
$18^{-50}\equiv-1$ (Mod 25) ..... (ii)
$\therefore$ $12^{50}+18^{50}\equiv-2$ (Mod 25)
$\equiv23$ (Mod 25)
$\therefore$ Answer = 23
Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.
Explanation:
$ \begin{aligned} & \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\ & \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\ & \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\ & \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\ & \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2) \end{aligned} $
From (1) and (2)
$ \Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8 $
$(1) \Rightarrow r=4$
so, coefficient of middle term is
$ { }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120 $
If the co-efficient of $x^9$ in ${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$ and the co-efficient of $x^{-9}$ in ${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$ are equal, then $(\alpha\beta)^2$ is equal to ___________.
Explanation:
$\because$ Both are equal
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
$\Rightarrow \frac{1}{\beta}=-\alpha$
$\Rightarrow \alpha \beta=-1$
$\Rightarrow(\alpha \beta)^{2}=1$
The remainder when (2023)$^{2023}$ is divided by 35 is __________.
Explanation:
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7
The constant term in the expansion of ${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$ is ___________.
Explanation:
$ \begin{aligned} & \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5} \end{aligned} $
Term independent of $x=$ coefficient of $x^{35}$ in
$ \begin{aligned} & ^{5} C_{4}\left(x^{8}(3 x+2)\right)^{4} \\\\ = & { }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4} \\\\ = & { }^{5} C_{4} \times{ }^{4} C_{3}(2)^{1}(3)^{3} \\\\ = & 5 \times 4 \times 2 \times 27 \\\\ = & 1080 \end{aligned} $
Let the sum of the coefficients of the first three terms in the expansion of ${\left( {x - {3 \over {{x^2}}}} \right)^n},x \ne 0.~n \in \mathbb{N}$, be 376. Then the coefficient of $x^4$ is __________.
Explanation:
$ \begin{aligned} & 3 n^{2}-5 n-250=0 \\\\ & n=10, \frac{-25}{3} \text { (Rejected) } \\\\ & T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\ & ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\ & ={ }^{10} C_{r} x^{10-3 r}(-3)^{r} \end{aligned} $
Here $r=2$
$ \begin{aligned} \text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\ & =45 \times 9 \\\\ & =405 \end{aligned} $
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}$
$\therefore$ $\alpha = {1012}$
$\sum\limits_{r=1}^{20}\left(r^{2}+1\right)(r !)$ is equal to
The remainder when $7^{2022}+3^{2022}$ is divided by 5 is :
The remainder when $(2021)^{2022}+(2022)^{2021}$ is divided by 7 is
$\sum\limits_{\matrix{ {i,j = 0} \cr {i \ne j} \cr } }^n {{}^n{C_i}\,{}^n{C_j}} $ is equal to
The remainder when $(11)^{1011}+(1011)^{11}$ is divided by 9 is
For two positive real numbers a and b such that ${1 \over {{a^2}}} + {1 \over {{b^3}}} = 4$, then minimum value of the constant term in the expansion of ${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$ is :
Let n $\ge$ 5 be an integer. If 9n $-$ 8n $-$ 1 = 64$\alpha$ and 6n $-$ 5n $-$ 1 = 25$\beta$, then $\alpha$ $-$ $\beta$ is equal to
If the constant term in the expansion of
${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$ is 2k.l, where l is an odd integer, then the value of k is equal to:
The term independent of x in the expansion of
$(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}},\,x \ne 0$ is :
If
$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $,
where $\alpha$ $\in$ R, then the value of 16$\alpha$ is equal to
The remainder when (2021)2023 is divided by 7 is :
The coefficient of x101 in the expression ${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}} + \,\,.....\,\, + \,\,{x^{500}}$, x > 0, is
If ${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$, then the remainder when K is divided by 6 is :
The remainder when 32022 is divided by 5 is :
$ \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.
expansion of (21/3 + 31/4)12 is :
expansion of ${\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}}$ is ${{10!} \over {{{(5!)}^2}}}$, then the value of 'a' is equal to :
than ${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$ is ______________.
(1 $-$ x)101 (x2 + x + 1)100 is :