Binomial Theorem

If ${I_1} + f = {(8\sqrt 3 + 13)^{13}},f' = {(8\sqrt 3 - 13)^{13}}$

${I_1} + f - f'=$ Even

${I_1} = $ Even

${I_2} + f - f' = {(7\sqrt 2 + 9)^9} + {(7\sqrt 2 - 9)^9}$

= Even

${I_2} = $ Even

For $\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$

${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$

$\therefore$ ${x^{15}} \to 3(15 - r) - {r \over 3} = 15$

$ \Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$

Similarly, for ${\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}$

${T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}$

$\therefore$ For ${x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6$

$\therefore$ ${}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1$

The coefficient of ${x^{301}}$ in

${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$

${}^{500}{C_{301}} + {}^{499}{C_{300}} + {}^{498}{C_{299}}\, + \,...\, + \,{}^{199}{C_0}$

$ = {}^{500}{C_{199}} + {}^{499}{C_{199}} + {}^{498}{C_{199}}\, + \,...\, + \,{}^{199}{C_{199}}$

$ = {}^{501}{C_{200}}$

$K = {2^{98}}$

$a = {}^{200}{C_{100}}\,{2^{50}}$

$\therefore$ ${{{}^{200}{C_{99}}\,.\,{2^{98}}} \over {{}^{200}{C_{100}}\,.\,{2^{50}}}} = {{{2^l}m} \over n}$

$ \Rightarrow {{100} \over {101}}\,.\,{2^{48}} = {{{2^l}m} \over n}$

$ \Rightarrow {{25} \over {101}}\,.\,{2^{50}} = {{{2^l}m} \over n}$

$\therefore$ $l = 50,m = 25,n = 101$

Given,

$1\,.\,{\left( {{}^{30}{C_1}} \right)^2} + 2\,.\,{\left( {{}^{30}{C_2}} \right)^2} + 3\,.\,{\left( {{}^{30}{C_3}} \right)^2}\, + \,.....\, + \,30\,.\,{\left( {{}^{30}{C_{30}}} \right)^2}$

$ = \sum\limits_{r = 1}^{30} {r\,.\,{{\left( {{}^{30}{C_r}} \right)}^2}} $

$ = \sum\limits_{r = 1}^{30} {r\,.\,{}^{30}{C_r}\,.\,{}^{30}{C_r}} $

$ = \sum\limits_{r = 1}^{30} {r\,.\,{{30} \over r}\,.\,{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $

$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}} $

$ = 30 \times $ (Coefficient of ${x^{29}}$ in ${(1 + x)^{59}}$)

$ = 30 \times \left( {{}^{59}{C_{29}}} \right)$

$ = 30 \times {{60} \over {30}} \times {}^{59}{C_{29}} \times {{30} \over {60}}$

$ = 30 \times {}^{60}{C_{30}} \times {{30} \over {60}}$

$ = 15 \times {{60!} \over {30!\,30!}}$

$\therefore$ $\alpha = 15$

$\sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_r}} $

$ = \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,{}^{23}{C_{23 - r}}} $ [using ${}^n{C_r} = {}^n{C_{n - r}}$]

$ = {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}{}^{23}{C_2} + {}^{22}{C_{22}}{}^{23}{C_1}$

We know,

${(1 + x)^{22}} = {}^{22}{C_0} + {}^{22}{C_1}x + {}^{22}{C_2}{x^2} + {}^{22}{C_3}{x^3}\, + \,...\, + \,{}^{22}{C_{22}}{x^{22}}$

and

${(1 + x)^{23}} = {}^{23}{C_0} + {}^{23}{C_1}x + {}^{23}{C_2}{x^2}\, + \,...\, + \,{}^{23}{C_{22}}{x^{22}} + {}^{23}{C_{23}}{x^{23}}$

Now coefficient of ${x^{23}}$ in ${(1 + x)^{22}}{(1 + x)^{23}}$ or ${(1 + x)^{45}}$

$ = {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}\,.\,{}^{23}{C_2} + {}^{22}{C_{22}}\,.\,{}^{23}{C_1}$

$ = \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_{23 - r}}} $

$\therefore$ Coefficient of ${x^{23}}$ in ${(1 + x)^{45}} = {}^{45}{C_{23}}$

Given,

$\sum\limits_{r = 1}^{20} {({r^2} + 1)(r!)} $

Let, $f(r) = ({r^2} + 1)(r!)$

$ = ({r^2})(r!) + r!$

$ = r(r\,r!) + r!$

$ = r[(r + 1 - 1)r!] + r!$

$ = r[(r + 1)r! - r!] + r!$

$ = r[(r + 1)! - (r!)] + r!$

$ = r(r + 1)! - r(r!) + r!$

$ = (r + 2 - 2)(r + 1)! - r(r!) + r!$

$ = (r + 2)(r + 1)! - 2(r + 1)! - [(r + 1 - 1)(r!)] + r!$

$ = (r + 2)! - 2(r + 1)! - (r + 1)! + r! + r!$

$ = (r + 2)! - 3(r + 1)! + 2r!$

$ = [(r + 2)! - (r + 1)!] - 2[(r + 1)! - r!]$

$\therefore$ $\sum\limits_{r = 1}^{20} {f(r)} $

$ = \sum\limits_{r = 1}^{20} {[(r + 2)! - (r + 1)!] - 2\sum\limits_{r = 1}^{20} {[(r + 1)! - r!]} } $

$ = [(22! + 21! + 20!\, + \,.....\, + \,4! + 3!) - (21! + 20! + 19!\, + \,....\, + \,3! + 2!] - 2[(21! + 20!\, + \,.....\, + \,3! + 2!) - (20! + 19!\, + \,.....\,1!)]$

$ = [(22!) - (2!)] - 2[(21)! - (1!)]$

$ = 22! - 2! - 2\,.\,(21)! + 2\,.\,1!$

$ = 22! - 2\,.\,(21)!$

${(2021)^{2022}} + {(2022)^{2021}}$

$ = {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}$

$ = {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1$

$ = {(7{k_2} - 1)^{674}} + (7m - 1)$

$ = (7n + 1) + (7m - 1) = 7(m + n)$ (multiple of 7)

$\therefore$ Remainder = 0

$\sum\limits_{i,\,j = 0\,\,i \ne j}^n {{}^n{C_i}\,{}^n{C_j} = \sum\limits_{i,\,j = 0}^n {{}^n{C_i}\,{}^n{C_j} - \sum\limits_{i = j}^n {{}^n{C_i}\,{}^n{C_j}} } } $

$ = \sum\limits_{j = 0}^n {{}^n{C_i}\,\sum\limits_{j = 0}^n {{}^n{C_j} - \sum\limits_{i = 0}^n {{}^n{C_i}\,{C_i}} } } $

$ = {2^n}\,.\,{2^n} - {}^{2n}{C_n}$

$ = {2^{2n}} - {}^{2n}{C_n}$

${\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)$

For ${\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)$

${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}$

So, remainder is 8

and ${\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0$

So, remainder is 8

Given, Binomial expansion,

${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$

General term,

${T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{10 - r} \over 8}}}\,.\,{x^{ - {r \over {12}}}}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{\left( {{{10 - r} \over 8} - {r \over {12}}} \right)}}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 3r - 2r} \over {24}}}}$

$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 5r} \over {24}}}}$

For constant term,

${{30 - 5r} \over {24}} = 0$

$ \Rightarrow r = 6$

$\therefore$ Constant term,

${T_{r + 1}} = {T_{6 + 1}} = {}^{10}{C_6}\,.\,{a^4}\,.\,{b^6}$

$ = {{10!} \over {6!\,4!}}{a^4}\,.\,{b^6}$

$ = {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}\,.\,{a^4}\,.\,{b^6}$

$ = 210{a^4}{b^6}$

We know, $GM \ge HM$

For terms a² and b³,

$\sqrt {{a^2}{b^3}} \ge {2 \over {{1 \over {{a^2}}} + {1 \over {{b^3}}}}}$

$ \Rightarrow \sqrt {{a^2}{b^3}} \ge {2 \over 4}$

$ \Rightarrow {a^2}{b^3} \ge {1 \over 4}$

$ \Rightarrow {({a^2}{b^3})^2} \ge {1 \over {16}}$

$\therefore$ ${a^4}{b^6} \ge {1 \over {16}}$

$\therefore$ Minimum value of ${a^4}{b^6} = {1 \over {16}}$

$\therefore$ Minimum value of constant term

${T_7} = 210 \times {a^4}{b^6}$

$ = 210 \times {1 \over {16}}$

$ = {{105} \over 8}$

Given,

${9^n} - 8n - 1 = 64\alpha $

$ \Rightarrow \alpha = {{{{(1 + 8)}^n} - 8n - 1} \over {64}}$

$ = {{\left( {{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,{8^1} + {}^n{C_2}\,.\,{8^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \right) - 8n - 1} \over {{8^2}}}$

$ = {{1 + 8n + {}^n{C_2}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^n} - 8n - 1} \over {{8^2}}}$

$ = {{{}^n{C_2}\,.\,{8^2} + {}^n{C_3}\,.\,{8^3}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \over {{8^2}}}$

$ = {}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,.....\,\,{}^n{C_n}\,.\,{8^{n - 2}}$

Also given,

${6^n} - 5n - 1 = 25\beta $

$ \Rightarrow \beta = {{{{(1 + 5)}^n} - 5n - 1} \over {25}}$

$ = {{{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,5 + {}^n{C_2}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^n} - 5n - 1} \over {{5^2}}}$

$ = {{1 + 5n + {}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^2} - 5n - 1} \over {{5^2}}}$

$ = {{{}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3} + {}^n{C_4}\,.\,{5^4}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^n}} \over {{5^2}}}$

$ = {}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}$

$\therefore$ $\alpha - \beta $

$ = \left( {{}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^{n - 2}}} \right) - \left( {{}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}} \right)$

$ = {}^n{C_3}\,.\,(8 - 5) + {}^n{C_4}\,.\,({8^2} - {5^2})\,\, + \,\,....\,\, + \,\,{}^n{C_n}({8^{n - 2}} - {5^{n - 2}})$

Given,

${\left( {3{x^2} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}$

$ = {{{{(3{x^8} - 2{x^7} + 5)}^{10}}} \over {{x^{50}}}}$

Now constant term in ${\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}} = {x^{50}}$ term in ${(3{x^8} - 2{x^7} + 5)^{10}}$

General term in ${(3{x^8} - 2{x^7} + 5)^{10}}$ is

$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3{x^8})^{{n_1}}}{( - 2{x^7})^{{n_2}}}{(5)^{{n_3}}}$

$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n^1}}}{( - 2)^{{n_2}}}{(5)^{{n^3}}}\,.\,{x^{8{n_1} + 7{n_2}}}$

$\therefore$ Coefficient of ${x^{8{n_1} + 7{n_2}}}$ is

$ = {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n_1}}}{( - 2)^{{n_2}}}{(5)^{{n_3}}}$

where ${n_1} + {n_2} + {n_3} = 0$

For coefficient of x⁵⁰ :

$8{n_1} + 7{n_2} = 50$

$\therefore$ Possible values of n₁, n₂ and n₃ are

$\therefore$ Coefficient of x⁵⁰

$ = {{10!} \over {1!\,6!\,3!}}{(3)^1}{( - 2)^6}{(5)^3}$

$ = {{10 \times 9 \times 8 \times 7} \over 6} \times 3 \times {5^3} \times {2^6}$

$ = 5 \times 3 \times 8 \times 7 \times 3 \times {5^3} \times {2^6}$

$ = 7 \times {5^4} \times {3^2} \times {2^9}$

$ = {2^k}\,.\,l$

$\therefore$ $l = 7 \times {5^4} \times {3^2}$ = An odd integer

and ${2^k} = {2^9}$

$ \Rightarrow k = 9$

General term of Binomial expansion ${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$ is

${T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}$

$ = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}$

In the term,

$\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$

Term independent of x is when

(1) $33 - 5r = 0$

$ \Rightarrow r = {{33} \over 5}\,\, \notin $ integer

(2) $33 - 5r = -2$

$ \Rightarrow 5r = 35$

$ \Rightarrow r = 7\,\, \in $ integer

(3) $33 - 5r = - 3$

$ \Rightarrow 5r = 36$

$ \Rightarrow r = {{36} \over 5}\,\, \notin $ integer

$\therefore$ Only for r = 7 independent of x term possible.

$\therefore$ Independent of x term

$ = - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)$

$ = - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)$

$ = {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}$

$ = {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}$

$ = {{33} \over {200}}$

Given,

$\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!)}}} } $

Now,

$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)} $

$ = \left( {{}^{31}{C_1}\,.\,{}^{31}{C_0} + {}^{31}{C_2}\,.\,{}^{31}{C_1} + {}^{31}{C_3}\,.\,{}^{31}{C_2} + \,\,......\,\, + \,\,{}^{31}{C_{31}}\,.\,{}^{31}{C_{30}}} \right)$

$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{31 - 1}} + {}^{31}{C_1}\,.\,{}^{31}{C_{31 - 2}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_{31 - 31}}} \right)$

[using ${}^n{C_r} = {}^n{C_{n - r}}$]

$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{30}} + {}^{31}{C_1}\,.\,{}^{31}{C_{29}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_0}} \right)$

$ = {}^{62}{C_{30}}$

Now, $\sum\limits_{k = 1}^{30} {{}^{30}{C_k}\,.\,{}^{30}{C_{k - 1}}} $

$ = \left( {{}^{30}{C_1}\,.\,{}^{30}{C_0} + {}^{30}{C_2}\,.\,{}^{30}{C_1} + \,\,.....\,\, + \,\,{}^{30}{C_{30}}\,.\,{}^{30}{C_{29}}} \right)$

$ = \left( {{}^{30}{C_0}\,.\,{}^{30}{C_{29}} + {}^{30}{C_1}\,.\,{}^{30}{C_{28}} + \,\,....\,\, + \,\,{}^{30}{C_{29}}\,.\,{}^{30}{C_0}} \right)$

$ = {}^{60}{C_{29}}$

$\therefore$ ${}^{60}{C_{30}} - {}^{60}{C_{29}} = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {29!\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {{{30!} \over {30}}\,.\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow {{60!} \over {30!\,31!}}\left( {{{62\,.\,61} \over {32}} - 30} \right) = {{\alpha (60!)} \over {30!\,31!}}$

$ \Rightarrow \alpha = {{62\,.\,61} \over {32}} - 30$

$ \Rightarrow 16\alpha = {{62\,.\,61 - 30 \times 32} \over 2}$

$ \Rightarrow 16\alpha = {{2822} \over 2} = 1411$

= (2016 + 5)²⁰²³ [here 2016 is divisible by 7]

= ²⁰²³C₀ (2016)²⁰²³ + .......... + ²⁰²³C₂₀₂₂ (2016) (5)²⁰²² + ²⁰²³C₂₀₂₃ (5)²⁰²³

= 2016 [²⁰²³C₀ . (2016)²⁰²² + ....... + ²⁰²³C₂₀₂₂ . (5)²⁰²²] + (5)²⁰²³

= 2016$\lambda$ + (5)²⁰²³

= 7 $\times$ 288$\lambda$ + (5)²⁰²³

= 7K + (5)²⁰²³ ...... (1)

Now, (5)²⁰²³

= (5)²⁰²² . 5

= (5³)⁶⁷⁴ . 5

= (125)⁶⁷⁴ . 5

= (126 $-$ 1)⁶⁷⁴ . 5

= 5[⁶⁷⁴C₀ (126)⁶⁷⁴ + ......... $-$ ⁶⁷⁴C₆₇₃ (126) + ⁶⁷⁴C₆₇₄]

= 5 $\times$ 126 [⁶⁷⁴C₀(126)⁶⁷³ + ....... $-$ ⁶⁷⁴C₆₇₃] + 5

= 5 . 7 . 18 [⁶⁷⁴C₀(126)⁶⁷³ + ....... $-$ ⁶⁷⁴C₆₇₃] + 5

= 7$\lambda$ + 5

Replacing (5)²⁰²³ in equation (1) with 7$\lambda$ + 5, we get,

(2021)²⁰²³ = 7K + 7$\lambda$ + 5

= 7(K + $\lambda$) + 5

$\therefore$ Remainer = 5

Given,

${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}}\,\, + $ ...... ${x^{500}}$

This is a G.P. with first term ${(5 + x)^{500}}$

Common ratio $ = {{x{{(5 + x)}^{499}}} \over {{{(5 + x)}^{500}}}} = {x \over {5 + x}}$ and 501 terms present.

$\therefore$ Sum $ = {{{{(5 + x)}^{500}}\left( {{{\left( {{x \over {5 + x}}} \right)}^{501}} - 1} \right)} \over {{x \over {5 + x}} - 1}}$

$ = {{{{(5 + x)}^{500}}\left( {{{{x^{501}} - {{(5 + x)}^{501}}} \over {{{(5 + x)}^{501}}}}} \right)} \over {{{x - 5 - x} \over {5 + x}}}}$

$ = {{{{{x^{501}} - {{(5 + x)}^{501}}} \over {5 + x}}} \over {{{ - 5} \over {5 + x}}}}$

$ = {1 \over 5}\left( {{{\left( {5 + x} \right)}^{501}} - {x^{501}}} \right)$

Coefficient of x¹⁰¹ in ${(5 + x)^{501}}$ is $ = {}^{501}{C_{101}}\,.\,{5^{400}}$

$\therefore$ In ${1 \over 5}\left( {{{(5 + x)}^{500}} - {x^{501}}} \right)$ coefficient of x¹⁰¹ is $ = {1 \over 5}\,.\,{}^{501}{C_{101}}\,.\,{5^{400}}$

$ = {}^{501}{C_{101}}\,.\,{5^{399}}$

${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$

$ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$

$ = {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}$

Now $K = {(1 + 2)^{10}} - {2^{10}}$

$ = {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}$

$ = {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}$

$ = 1 + 20 + 5120 + 6\lambda $

$ = 5136 + 6\lambda + 5$

$ = 6\mu + 5$

$\lambda ,\,\mu \in N$

$\therefore$ remainder = 5

${3^{2022}}$

$ = {({3^2})^{1011}}$

$ = {(9)^{1011}}$

$ = {(10 - 1)^{1011}}$

$ = {}^{1011}{C_0}{(10)^{1011}} + \,\,.....\,\, + \,\,{}^{1011}{C_{1010}}\,.\,{(10)^1} - {}^{1011}{C_{1011}}$

$ = 10\left[ {{}^{1011}{C_0}{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right] - 1$

$ = 10\,K - 1$

[As $10\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right]$ is multiple of 10]

$ = 10K + 5 - 5 - 1$

$ = 10K - 5 + 5 - 1$

$ = 5(2K - 1) + 4$

$\therefore$ Unit digit = 4 when divided by 5.