Binomial Theorem

${T_5} = {}^8{C_4}{{{x^{12}}} \over {81}} \times {{81} \over {{x^4}}} = 5670$

$ \Rightarrow 70{x^8} = 5670$

$ \Rightarrow x = \pm \sqrt 3 $

Given sum = coefficient of x^r in the expansion of

(1 + x)²⁰(1 + x)²⁰,

Which is equal to ⁴⁰C_r

It is maximum when r = 20

The general term in the expansion of the binomial expression $(a+b)^n$ is

$ T_{r+1}={ }^n C_r a^{n-r} b^r $

Therefore, the general term in the expansion of the binomial expression

$x^2\left(\sqrt{x}+\frac{\lambda}{x^2}\right)^{10}$ is

$ \begin{aligned} T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\\\ & ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\\\ & ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r} \end{aligned} $

Now, for the coefficient of $x^2$,

$ \begin{aligned} 2+\frac{10-r}{2}-2 r =2 \\\\ \Rightarrow \frac{10-r}{2}-2 r =0 \\\\ \Rightarrow 10-r =4 r \Rightarrow r=2 \end{aligned} $

So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$

$ \begin{aligned} \Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\\\ \Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\\\ \Rightarrow & 45 \lambda^2 =720 \\\\ \Rightarrow & \lambda^2 =16 \\\\ \Rightarrow & \lambda = \pm 4 \end{aligned} $

Given the problem asks for the positive value of $\lambda$, $\lambda = 4$.

So, the correct option is (A) 4.

${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$

$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$

$ \Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$

$ \Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$

$ \Rightarrow 100 = k$

${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$

${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$

$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$

$ \Rightarrow \,\,{x^{2\log 2x}} = 256$

$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$

$ \Rightarrow 2{({\log _2}x)^2} = 8$

$ \Rightarrow \,\,{({\log _2}x)^2} = 4$

$ \Rightarrow \,\,{\log _2}x = 2$  or  $-$ 2

$x = 4$   or  ${1 \over 4}$

${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$

= (1 $-$ t⁶)³ (1 $-$ t)^$-$3

= (1 $-$ ³C₁t⁶ + ³C₂t¹² $-$ ³C₃t¹⁸) $ \times $ (1 $-$ t)^$-$3

coefficient of t⁴ is 1 $ \times $ coefficient of t⁴ in (1 $-$ t)^$-$3

= 1 $ \times $ ^3+4$-$1C₄ (By multinomial theorem)

= ⁶C₄ = 15

${{{2^{403}}} \over {15}}$

$ = {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$

$ = {8 \over {15}}{\left( {16} \right)^{100}}$

$ = {8 \over {15}}{\left( {15 + 1} \right)^{100}}$

$ = {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$

$ = {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$

$ = {8 \over {15}} + 8$ (integer)

$ \therefore $  Fractional part $ = {8 \over {15}}$

According to the question,

${k \over {15}} = {8 \over {15}}$

$ \Rightarrow $   K $=$ 8

Given,

(2 $-$ x²) . (1 + 2x + 3x²) ⁶ + (1 $-$ 4x²)⁶)

Let, a = ((1 + 2x + 3x²)⁶ + (1 $-$ 4x²)⁶)

$\therefore\,\,\,\,$ Given statement becomes,

(2 $-$ x²) . (a)

=    2a $-$ x² (a)

Here coefficients of x² is

=    2 (coefficient of x² in a ) $-$ 1 (constant in a)

(1 + 2x + 3x²) ⁶  =   ⁶C₀ + ⁶C₁ (2x + 3x²) + ⁶C₂ (2x + 3x²)²

      + . . . . . .+ (2x + 3x²)⁶

(1 $-$ 4x²)⁶ = ⁶C₀ $-$ ⁶C₁ (4x²) + ⁶C₂ (4x²)²

      + . . . . .+ (4x²)⁶

Coefficient of x² in (1 + 2x + 3x²)⁶

= ⁶C₁ $ \times $ 3 + ⁶C₂ $ \times $ 4

= 18 + 60

Coefficient of x² in (1 $-$ 4x²)⁶

= $-$ ⁶C₁ $ \times $ 4

= $-$ 24

Coefficient of x² in ((1 + 2x + 3x²)⁶ + (1 $-$ 4x²)⁶)

= 60 + 18 $-$ 24

= 54

Constant term in (1 + 2x + 3x²)⁶ = ⁶C₀ = 1

Constant term in (1 $-$ 4x)⁶ = ⁶C₀ = 1

$\therefore\,\,\,\,$ Constant term in ((1 + 2x + 3x²)⁶ + (1 $-$ 4x)⁶)

= 1 + 1 = 2

$\therefore\,\,\,\,$ Coefficient of x² in (2 $-$ x²) ((1 + 2x + 3x²)⁶ + (1 $-$ 4x²)⁶)

= 2 (54) $-$ 1 (2)

= 108 $-$ 2

= 106

$ \because $$\,\,\,$ (1 + x)² = 1 + 2x + x²,

(1 + x²)³ = 1 + 3x² + 3x⁴ + x⁶

and (1 + x³)⁴ = 1 + 4x³ + 6x⁶ + 4x⁹ + x¹²

So, the possible combination for x¹⁰ are :

x . x⁹,   x . x⁶.  x³,  x² . x² . x⁶ , x⁴ . x⁶

Corresponding coefficients are 2 $ \times $ 4, 2$ \times $ 1 $ \times $4, 1 $ \times $3$ \times $ 6,

3 $ \times $ 6   or   8, 8, 18, 18.

$\therefore\,\,\,$ Sum of the coefficient is 8 + 8 + 18 + 18 = 52

Therefore, the coefficient of x¹⁰ in the expansion of

(1 + x)² (1 + x²)³ (1 + x³)⁴ is 52.

Given,
${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$ + ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$

= ${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$

+ ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$

= ${\left[ {{{2(\sqrt {5x^3 + 1} + \sqrt {5x^3 - 1} } \over 2}} \right]^8}$ + ${\left[ {{{2(\sqrt {5x^3 - 1} - \sqrt {5x^3 - 1} } \over 2}} \right]^8}$

= (${\sqrt {5{x^3} + 1} }$ + ${\sqrt {5{x^3} - 1} }$)⁸ + (${\sqrt {5{x^3} - 1} }$ - ${\sqrt {5{x^3} - 1} }$)⁸

= 2 $\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}$ +

      + ${}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1} $

      + ${}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}$

      + ${}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}$

      + $\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]$

= 2$[{}^8{C_0}{(5{x^3} + 1)^4}$

   + ${}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)$

   + ${}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}$

   + ${}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}$

   + ${}^8{C_8}{(5{x^3} - 1)^4}]$

Here maximum power of x is 12

$ \therefore $ Degree of polynomial = 12

Coefficient of x¹²

= 2 [⁸C₀ 5⁴ + ⁸C₂ $\times$ 5³ $\times$ 5 + ⁸C₄ $\times$ 5² $\times$ 5² + ⁸C₆ $\times$ 5 $\times$ 5³ + ⁸C₈ $\times$ 5⁴]

= 160000 = (20)⁴

${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$

= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$

[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 = -5

then $r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$ = ${{25} \over 3} \times {6 \over 5}$ = 10

$\therefore$ T₁₁ is the term with x^-5.

$\therefore$ T₁₁ = ${}^{10}{C_{10}}$ = 1

We have,

${{{{\left( {27} \right)}^{999}}} \over 7}$

= ${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$

= ${{28\,\lambda - 1} \over 7}$

= ${{28\,\lambda - 7 + 7 - 1} \over \lambda }$

= ${{7\left( {4\lambda - 1} \right) + 6} \over 7}$

$\therefore\,\,\,$ Remainder = 6

T_r+1  =  ¹⁸C_r  ${\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}$  .  ${\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}$

=  ¹⁸C_r  ${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}$

For coefficient of x^$-$2,

${{18 - 2r} \over 3}$   =   $-$2

$ \Rightarrow $   r  =  12

$ \therefore $   Coefficient of   x^$-$2     is  (m) = ¹⁸C₁₂  ${\left( {{1 \over 2}} \right)^{12}}$

For coefficient of x^$-$4,

${{18 - 2r} \over 3}$ = $-$ 4

$ \Rightarrow $   r = 15

$ \therefore $   Coefficient of x^$-$4 is (n) = ¹⁸C₁₅ $\left( {{1 \over {2}}} \right)$¹⁵

$ \therefore $   ${m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}$

= ${{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}$

= 182

Assume,

P = (1 + x)²⁰¹⁶ + x(1 + x)²⁰¹⁵ + . . . . .+ x²⁰¹⁵ . (1 + x) + x²⁰¹⁶    . . . . .(1)

Multiply this with $\left( {{x \over {1 + x}}} \right),$

$\left( {{x \over {1 + x}}} \right)P = $ x(1 + x)²⁰¹⁵ + x²(1 + x)²⁰¹⁴ +

        . . . . . . + x²⁰¹⁶ + ${{{x^{2017}}} \over {1 + x}}$ . . . . . (2)

Performing (1) $-$ (2), we get

${P \over {1 + x}} = $ (1 + x)²⁰¹⁶ $-$ ${{{x^{2017}}} \over {1 + x}}$

$ \Rightarrow $    P = (1 + x)²⁰¹⁷ $-$ x²⁰¹⁷

$ \therefore $    a₁₇ = coefficient of x¹⁷ $=$ ²⁰¹⁷C₁₇ $=$ ${{2017!} \over {17!\,\,2000!}}$

Total no of terms in ${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}$ = ${}^{n + 2}{C_2}$ = 28

(n+2)(n+1) = 56

$ \Rightarrow n = 6$

Sum of coefficient = (1 - 2 + 4)⁶ = 3⁶ = 729

${\left( {1 - 2\sqrt x } \right)^{50}}$

= ${}^{50}{C_0} + {}^{50}{C_1}.\left( { - 2\sqrt x } \right) + {}^{50}{C_2}.{\left( { - 2\sqrt x } \right)^2} + ....$

Now we need to find out those coefficient where degree of x is integer and you can see at odd terms power of x is integer.

Let ${\left( {1 - 2\sqrt x } \right)^{50}}$ = Odd(A) - Even(B)

So ${\left( {1 + 2\sqrt x } \right)^{50}}$ = A + B

$\therefore$ 2A = ${\left( {1 + 2\sqrt x } \right)^{50}}$ + ${\left( {1 - 2\sqrt x } \right)^{50}}$

$ \Rightarrow A = {1 \over 2}\left[ {{{\left( {1 + 2\sqrt x } \right)}^{50}} + {{\left( {1 - 2\sqrt x } \right)}^{50}}} \right]$

Now to find sum of coefficient of A, put x = 1.

$\therefore$ Sum of coefficient of A = ${1 \over 2}\left[ {{{\left( {1 + 2} \right)}^{50}} + {{\left( {1 - 2} \right)}^{50}}} \right]$

= ${1 \over 2}\left[ {{{\left( 3 \right)}^{50}} + 1} \right]$

$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$

= ${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$

= $\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$

Coefficient of x³ is
${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$ = 0

$ \Rightarrow 153a - 9b = 1632$ ....... (1)

Coefficient of x⁴ is
${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$ = 0

$ \Rightarrow 3b - 32a = -240$ ....... (2)

Solving (1) and (2), we get $a$ = 16, b = ${{172} \over 3}$

${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$

= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$

[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

Let ${\left( {a + x} \right)^n}$ = Odd trems(A) + Even terms(B)

So ${\left( {a - x} \right)^n}$ = Odd terms(A) - Even terms(B)

$\therefore$ ${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$

= (A + B) - (A - B)

= 2B

= 2[even terms]

= 2[ T₂ + T₄ + T₆ + ....... ]

So in case of ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$

= 2[ T₂ + T₄ + T₆ + ....... ]

= 2[ ${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$

Here in ${\left( {\sqrt 3 } \right)^{2n - 1}}$, 2n - 1 is odd number. So there will be always ${\sqrt 3 }$ in ${\left( {\sqrt 3 } \right)^{2n - 1}}$.

So ${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$ will be always irrational number.

Given,
${\left( {1 - x - {x^2} + {x^3}} \right)^6}$

= ${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$

= ${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$

= $\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times$
        $\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)$

$\therefore$ Coefficient of x⁷ = $ - {}^6{C_1} \times - {}^6{C_3} + \left( { - {}^6{C_3}} \right) \times {}^6{C_2} + \left( { - {}^6{C_5}} \right) \times - {}^6{C_1}$

= 120 - 300 + 36 = - 144

Note :

$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $ = $ = n{.2^{n - 1}}$

$\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}} = n\left( {n + 1} \right){2^{n - 2}}$

Given that,

${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$

=$\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}$

= 10$ \times $11$ \times $${2^{10 - 2}}$ - 10$ \times $${2^{10 - 1}}$

= 10$ \times $${2^{8}}$(11 - 2)

= 10$ \times $9$ \times $${2^{8}}$

= 90$ \times $${2^{8}}$

${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$

= 10$ \times $${2^{10-1}}$

= 10$ \times $${2^{9}}$

${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$

= 10$ \times $11$ \times $${2^{10-2}}$

= ${{110} \over 2} \times {2^9}$

= 55 $ \times $ $ {2^9}$

${8^{2n}} - {\left( {62} \right)^{2n + 1}}$

= ${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$

= ${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$

= $\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$
         + $\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)$

= 2 + 63$\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]$

= 63$ \times $[Some integral value] + 2

63$ \times $[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.

Check Statement - 1

$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}}$

= $\sum\limits_{r = 0}^n {r.{}^n{C_r}} $ + $\sum\limits_{r = 0}^n {{}^n{C_r}} $

= $\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}} $ $ + {2^n}$

= $n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} $ $ + {2^n}$

= $n \times {2^{n - 1}}$$ + {2^n}$

= ${2^{n - 1}}\left[ {n + 2} \right]$

Check Statement 2 :

$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}$

= $\sum\limits_{r = 0}^n r .{}^n{C_r}.{x^r}$ + $\sum\limits_{r = 0}^n {{}^n{C_r}.{x^r}} $

= $n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^r}} $ $ + {\left( {1 + x} \right)^n}$

= $nx\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^{r - 1}}} + {\left( {1 + x} \right)^n}$

= $nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n}$

Substitude x = 1 in the statement 2 and we get,

$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $

So Option B is correct.

We know

${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0$

$ \Rightarrow $ $({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$ $+{}^{20}{C_{10}}$ $(-{}^{20}{C_{9}}+ {}^{20}{C_{8}}+...... + {}^{20}{C_{0}})$

(As ${}^{20}{C_{11}} = {}^{20}{C_9}$)

$ \Rightarrow $ $2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$ $+{}^{20}{C_{10}}$ = 0

$ \Rightarrow $ ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$ = $ - {1 \over 2}{}^{20}{C_{10}}$

Adding ${}^{20}{C_{10}}$ both sides,

$ \Rightarrow $ ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$ $ + {}^{20}{C_{10}}$ = $ - {1 \over 2}{}^{20}{C_{10}}$ $ + {}^{20}{C_{10}}$

$ \Rightarrow $ ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$ $ + {}^{20}{C_{10}}$ = $ {1 \over 2}{}^{20}{C_{10}}$

According to the question,

t₅ + t₆ = 0

$\therefore$ ${}^n{C_4}.{a^{n - 4}}.{b^4}$ + $\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)$ = 0

By solving we get,

${a \over b} = {{n - 4} \over 5}$

${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,$

= $\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)$ -
                $\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)$

${a_1}$ = Coefficient of y = ${{}^n{C_1}}$ - ${{}^m{C_1}}$ = 10

$ \Rightarrow $ n - m = 10

${a_2}$ = Coefficient of y²

= ${}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10$

$ \Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10$

$ \Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20$

$ \Rightarrow$ (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20

$ \Rightarrow$ 90 + 19m + m² - 2m² - 20m + m² - m - 20 = 0

$ \Rightarrow$ 70 - 2m = 0

$ \Rightarrow$ m = 35

$\therefore$ n = 10 + 35 = 45

${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$

= ${\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}$

= $\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$ -
               $\left[ {1 + \left( { - 1} \right)\left( { - bx} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - bx} \right)}^2} + ...} \right]$

= $\left[ {1 + ax + {a^2}{x^2} + ... + {a^{n - 1}}{x^{n - 1}} + {a^n}{x^n}}+.... \right]$ -
               $\left[ {1 + bx + {b^2}{x^2} + ... + {b^{n - 1}}{x^{n - 1}} + {b^n}{x^n}}+.... \right]$

Coefficient of xⁿ =

${a^n} + {a^{n - 1}}b + {a^{n - 2}}{b^2} + .... + {b^n}$

= ${a^n}\left[ {1 + {b \over a} + {{{b^2}} \over {{a^2}}} + ..... + {{{b^n}} \over {{a^n}}}} \right]$

= ${a^n}\left[ {{{{{\left( {{b \over a}} \right)}^{n + 1}} - 1} \over {{b \over a} - 1}}} \right]$

= ${a^n}\left[ {{{{b^{n + 1}} - {a^{n + 1}}} \over {{a^{n + 1}}\left( {{{b - a} \over a}} \right)}}} \right]$

= ${{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}}$

Given, ${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}} $

$ \Rightarrow $ ${}^{50}{C_4}$ + ${}^{55}{C_3}$ + ${}^{54}{C_3}$ + ${}^{53}{C_3}$ + ${}^{52}{C_3}$ + ${}^{51}{C_3}$ + ${}^{50}{C_3}$

Arrange those this way

$ \Rightarrow $ ${}^{50}{C_4}$ + ${}^{50}{C_3}$ + ${}^{51}{C_3}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

We know this formula [ ${{}^n{C_r}}$ + ${{}^n{C_{r - 1}}}$ = ${{}^{n + 1}{C_r}}$ ] which is used to solve this problem.

$ \Rightarrow $ ${}^{51}{C_4}$ + ${}^{51}{C_3}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$ \Rightarrow $ ${}^{52}{C_4}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$ \Rightarrow $${}^{53}{C_4}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$ \Rightarrow$$ {}^{54}{C_4}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$ \Rightarrow $${}^{55}{C_4}$ + ${}^{55}{C_3}$

$ \Rightarrow$$ {}^{56}{C_4}$

Let r = 2

$\therefore$ 2nd, 3rd and 4th terms are in AP.

2nd term = T₂ = ${}^m{C_1}.y$

Coefficient of T₂ = ${}^m{C_1}$

3rd term = T₃ = ${}^m{C_2}.{y^2}$

Coefficient of T₃ = ${}^m{C_2}$

4th term = T₄ = ${}^m{C_3}.{y^3}$

Coefficient of T₂ = ${}^m{C_3}$

$\therefore$ 2.${}^m{C_2}$ = ${}^m{C_1}$ + ${}^m{C_3}$

$ \Rightarrow $ $2.{{m\left( {m - 1} \right)} \over {1.2}}$ = ${m \over 1}$ + ${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$

$ \Rightarrow $ 6m² - 6m = 6m +m(m² - 3m + 2)

$ \Rightarrow $ 6m² - 6m = 6m + m³ - 3m² + 2m

$ \Rightarrow $ 6m - 6 = 6 + m² - 3m + 2

$ \Rightarrow $ m² - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, ${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$ putting r = 2 we get

m² - 9m + 14 = 0. So Option C is correct.

${\left( {1 + x} \right)^{{3 \over 2}}}$ = 1 + ${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$

= 1 + ${3 \over 2}x + {3 \over 8}{x^2}$ (As $x$ is so small, so ${x^3}$ and higher powers of $x$ neglected)

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

= ${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$

= ${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$

= ${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$

= ${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$

= $ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$

[ As x³ is so small we can ignore $-{3 \over {16}}{x^3}$]

= $ - {3 \over 8}{x^2}$

General term of ${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$ is T_r+1.

T_r+1 = ${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$

= ${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$

For the coefficient of x⁷,

$ \Rightarrow $ 22 - 3r = 7

$ \Rightarrow $ r = 5

So coefficient of x⁷ = ${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$ ......(1)

Now General term of ${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$ is T_r+1.

T_r+1 = ${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$

= ${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$

For the coefficient of x^-7,

11 - 3r = -7

$ \Rightarrow $ r = 6

$\therefore$ Coefficient of x^-7 = ${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$

According to question,

Coefficient of x⁷ = Coefficient of x^-7

$ \Rightarrow $ ${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$ = ${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$

$ \Rightarrow $ $ab = 1$

${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}$

=${1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}$

${t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}$

= ${0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}$.........(1)

We can write ${t_n}$ by rearranging like this,

${t_n}$ = ${n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}$

= ${n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}$.........(2)

[as ${{}^n{C_0}}$ = ${{}^n{C_n}}$, ${{}^n{C_1}}$ = ${{}^n{C_{n - 1}}}$......]

By adding (1) and (2) we get,

$2{t_n}$ = ${n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}$

= $n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]$

= n${S_n}$

$\therefore$ ${{{t_n}} \over {{S_n}}} = {n \over 2}$

Given $\left( {1 + x} \right){\left( {1 - x} \right)^n}$

= ${\left( {1 - x} \right)^n}$ + $x{\left( {1 - x} \right)^n}$

General term of ${\left( {1 - x} \right)^n}$ = ${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$

$\therefore$ Term containing ${x^n}$ in ${\left( {1 - x} \right)^n}$ = ${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$

So coefficient of ${x^n}$ in ${\left( {1 - x} \right)^n}$ = ${}^n{C_n}.{\left( { - 1} \right)^n}$

General term of $x{\left( {1 - x} \right)^n}$ = ${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$

$\therefore$ Term containing ${x^n}$ in $x{\left( {1 - x} \right)^n}$ = ${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$

So coefficient of ${x^n}$ in $x{\left( {1 - x} \right)^n}$ = ${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$

$\therefore$ coefficient of ${x^n}$ in ${\left( {1 - x} \right)^n}$ + $x{\left( {1 - x} \right)^n}$

= ${}^n{C_n}.{\left( { - 1} \right)^n}$ + ${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$

= ${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$

= ${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$

= ${\left( { - 1} \right)^n}\left[ {1 - n} \right]$

For ${\left( {1 + \alpha x} \right)^4}$ the middle term ${T_{{4 \over 2} + 1}}$ = ${}^4{C_2}.{\alpha ^2}{x^2}$

$\therefore$ Coefficient of middle term = ${}^4{C_2}.{\alpha ^2}$

For ${\left( {1 - \alpha x} \right)^6}$ the middle term ${T_{{6 \over 2} + 1}}$ = ${}^6{C_3}.-{\alpha ^3}{x^3}$

$\therefore$ Coefficient of middle term = ${}^6{C_3}.{-\alpha ^3}$

$\therefore$ According to question,

${}^4{C_2}.{\alpha ^2}$ = ${}^6{C_3}.{-\alpha ^3}$

$ \Rightarrow 6 = 20 \times - \alpha $

$ \Rightarrow \alpha = - {3 \over {10}}$

General term of ${\left( {1 + x} \right)^{n}}$ is (${T_{r + 1}}$) = ${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$

$\therefore$ General term of ${\left( {1 + x} \right)^{27/5}}$ = ${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}$

For first negative term, ${\left( {{{27} \over 5} - r + 1} \right)}$ < 0

$ \Rightarrow r > {{27} \over 5} + 1$

$ \Rightarrow r > {{32} \over 5}$

$ \Rightarrow r > 6.4$

$\therefore$ r = 7

${T_{7 + 1}} = {T_8}$ means 8^th term is the first negative term.

General term = ${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$
= ${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$

When ${{{256 - r} \over 2}}$ is integer then ${\left( 3 \right)^{{{256 - r} \over 2}}}$ is integer.
And when ${{r \over 8}}$ is integer then ${\left( 5 \right)^{{r \over 8}}}$ is integer.

Entire general term will be integer when ${{{256 - r} \over 2}}$ and ${{r \over 8}}$ both are integer.

${{{256 - r} \over 2}}$ is integer when r = 0, 2, 4, 6, ......, 256

${{r \over 8}}$ is integer when r = 0, 8, 16 ,......., 256

Now both ${{{256 - r} \over 2}}$ and ${{r \over 8}}$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$\therefore$ 256 = 0 + (n - 1)8 using formula of AP, t_n = a + (n - 1)d

$\therefore$ n = ${{256} \over 8} + 1$ = 32 + 1 = 33

${\left( {1 + 0.0001} \right)^{10000}}$ = ${\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}$

= 1 + 10000${ \times {1 \over {{{10}^4}}}}$ + ${{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}$+......$\infty $

< 1 + 1 + ${1 \over {2!}}$ + ${1 \over {3!}}$ + ...... $\infty $ = e = 2.71828 < 3

We know, $\,{\left( {a + b} \right)^n}$ = ${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$

Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable.

So put $a$ = b = 1

$\therefore$ 2ⁿ = ${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$

According to question, 2ⁿ = 4096 = 2¹²

$ \Rightarrow n = 12$

So $\,{\left( {a + b} \right)^n}$ = $\,{\left( {a + b} \right)^{12}}$

Here n = 12 is even so formula for greatest term is
${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$

For n = 12, greatest term ${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$

$\therefore$ Coefficient of the greatest term = ${}^{12}{C_6}$ = ${{12!} \over {6!6!}}$ = 924

$\,{\left( {r + 2} \right)^{th}}$ term = ${}^{2n}{C_{r+1}}{\left( x \right)^r}$

And coefficient of $\,{\left( {r + 2} \right)^{th}}$ = ${}^{2n}{C_{r+1}}$

$3{r^{th}}$ term = ${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$

And coefficient of $3{r^{th}}$ term = ${}^{2n}{C_{3r - 1}}$

According to the question,
${}^{2n}{C_{r+1}}$ = ${}^{2n}{C_{3r - 1}}$

$ \Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n$

[As if ${}^n{C_p} = {}^n{C_q}$ then p + q = n]

$ \Rightarrow 4r = 2n$

$ \Rightarrow n = 2r$

Here in this expansion ${\left( {1 + x} \right)^{p + q}}$

The general term = ${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$

$\therefore$ ${x^p}$ will be present in the term = ${}^{p + q}{C_p}.{\left( x \right)^p}$

So coefficient of ${x^p}$ = ${}^{p + q}{C_p}$

And ${x^q}$ will be present in the term = ${}^{p + q}{C_q}.{\left( x \right)^q}$

$\therefore$ coefficient of ${x^q}$ = ${}^{p + q}{C_q}$

We know ${}^n{C_r}$ = ${}^n{C_{n - r}}$

$\therefore$ ${}^{p + q}{C_q}$ = ${}^{p + q}{C_{\left( {p + q} \right) - q}}$ = ${}^{p + q}{C_p}$

So coefficients of ${x^p}$ and ${x^q}$ are equal.