Binomial Theorem

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}$

$\therefore$ $\alpha = {1012}$

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$

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.$

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$

$\beta = {{12} \over 5}$

$d = - 6\beta - {\beta ^2}$

$\therefore$ $50 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57$

$l = 1 + (1 + {}^{49}{C_0} + {}^{49}{C_1}\, + \,....\, + \,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}})$

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 $n = 98$

$m + n = 99$

${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$

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$

$\therefore$ $k + {n_0} = 24$

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$

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$

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

$r = 5$.

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.

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$

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$

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

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}$

$\therefore$ $\beta = 83$

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$

$\therefore$ $\alpha + \beta = 275 + 11 = 286$

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.