(1001)⁵ = (1000 + 1)⁵
Ans. 1005010010005001

\(\left(1+x^{2}+2 x\right)^{20}=\left\{(1+x)^{2}\right]^{20}=(1+x)^{40}\)
Ans.1

16y (4y²- 3)

Ans. 5

\(\frac{35}{48}x^6\)

3421440

2(5a⁴b + 10a²b³ + b⁵); 568\(\sqrt3\)

9039207968

\(\sum _{ r=0 }^{ n }{ { 3 }^{ r } } \) ⁿC_r 
= 3^o nC_o + 3^1 nC₁ + 3^2 nC₂ + .... + 3^n nC_n
= ⁿC_o (1)ⁿ . 3^o + ⁿC₁ (1)^n-1 . 3¹ + ⁿC₂(1)^n-2 . 3² + ...... + ⁿC_n 3ⁿ
= ( 1+ 3)ⁿ = 4ⁿ

n\(\pi +(-1)^{ n }\frac { \pi }{ 6 } \)

(x + 1)⁶ + (x - 1)⁶ 
= [⁶C_ox⁶ + ⁶C₁x⁵ + ⁶C₂x⁴ + ⁶C₃x³ + ⁶C₄x² + ⁶C₅x + ⁶C₆] + [⁶C_ox⁶ + ⁶C₁x⁵(-1) + ⁶C₂x⁴(-1)² + ⁶C₃x³(-1)³ + ⁶C₄x²(-1)⁴ + ⁶C₅x(-1)⁵+ ⁶C₆(-1)⁶]
= [x⁶ + 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1] - [x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1]
= 2x⁶ + 30x⁴ + 30x² + 2
= 2( x⁶ + 15x⁴ + 15x² + 1)
Putting x = \(\sqrt { 2 } \)
\((\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}\)
= 2 [​​​​​(​​\(\sqrt { 2 } \))⁶ + 15\(\left( \sqrt { 2 } \right) ^{ 4 }\) + 15\(\left( \sqrt { 2 } \right) ^{ 2 }\) + 1]
= 2 [8 + 15 x 4 + 15 x 2 + 1]
= 2 [ 8 + 60 + 30 + 1]
= 2 x 99 = 198

We have, (2³)ⁿ-7n-1=(8)ⁿ-7n-1
Here, 8 can be written as 1+7. 
\(\therefore { 2 }^{ 3n }-7n-1=\left( 1+7 \right) ^{ n }-7n-1\)
by using binomial theorem, we get
\({ 2 }^{ 3n }-7n-1=\left( ^{ n }{ C }_{ 0 }+^{ n }{ C }_{ 1 }7+^{ n }{ C }_{ 2 }{ 7 }^{ 2 }+...+^{ n }{ C }_{ n }{ 7 }^{ n } \right) -7n-1\)
\(=\left( 1+\frac { n! }{ 1!\left( n-1 \right) ! } .7+\frac { n! }{ 2!\left( n-2 \right) ! } .{ 7 }^{ 2 }+{ 7 }^{ n } \right) -7n-1\)
\(=1+7n+\frac { n\left( n-1 \right) }{ 2 } { 7 }^{ 2 }+...+{ 7 }^{ n }-7n-1\)
\(=\frac { n\left( n-1 \right) }{ 2 } { 7 }^{ 2 }+....+{ 7 }^{ n }\)
Take common number outside, we get
\({ 2 }^{ 3n }-7n-1={ 7 }^{ 2 }\left[ \frac { n\left( n-1 \right) }{ 2 } +...+{ 7 }^{ n-2 } \right] \)
\(=49\left[ \frac { n\left( n-1 \right) }{ 2 } +...+{ 7 }^{ n-2 } \right] \)
Here, we see that above series is a multiple of 49.
Therefore, 2³ⁿ -7n-1 is divisible by 49.

We have,\(\left( { x+1 } \right) ^{ 6 }=^{ 6 }{ C }_{ 0 }{ x }^{ 6 }+^{ 6 }{ C }_{ 1 }{ x }^{ 5 }\times 1+^{ 6 }{ C }_{ 2 }{ x }^{ 4 }\times (1)^{ 2 }+^{ 6 }{ C }_{ 3 }{ x }^{ 3 }\times (1)^{ 3 }+^{ 6 }{ C }_{ 4 }{ x }^{ 2 }\times (1)^{ 4 }+^{ 6 }{ C }_{ 5 }{ x }\times (1)^{ 5 }+^{ 6 }{ C }_{ 6 }(1)^{ 6 }\)\(\Rightarrow \left( { x+1 } \right) ^{ 6 }=^{ 6 }{ C }_{ 0 }{ x }^{ 6 }+^{ 6 }{ C }_{ 1 }{ x }^{ 5 }\times 1+^{ 6 }{ C }_{ 2 }{ x }^{ 4 }\times (1)^{ 2 }+^{ 6 }{ C }_{ 3 }{ x }^{ 3 }+^{ 6 }{ C }_{ 2 }{ x }^{ 2 }+^{ 6 }{ C }_{ 1 }{ x }+^{ 6 }{ C }_{ 0 }\quad [\because ^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }]\)
\(\Rightarrow \left( { x+1 } \right) ^{ 6 }={ x }^{ 6 }+{ 6x }^{ 5 }+\frac { 6\times 5 }{ 2 } { x }^{ 4 }+\frac { 6\times 5\times 4 }{ 6 } { x }^{ 3 }+\frac { 6\times 5 }{ 2 } { x }^{ 2 }+6x+1\)
\(\Rightarrow \left( { x+1 } \right) ^{ 6 }={ x }^{ 6 }+{ 6x }^{ 5 }+15{ x }^{ 4 }+20{ x }^{ 3 }+15{ x }^{ 2 }+6x+1...(i)\)
Similarly,
\(\left( { x-1 } \right) ^{ 6 }={ x }^{ 6 }-{ 6x }^{ 5 }+15{ x }^{ 4 }-20{ x }^{ 3 }+15{ x }^{ 2 }-6x+1...(ii)\)
On adding Eqs. (i) and (ii), we get
\(\left( { x+1 } \right) ^{ 6 }+\left( { x-1 } \right) ^{ 6 }=2[{ x }^{ 6 }+15{ x }^{ 4 }+15{ x }^{ 2 }+1]\)
Now , on putting \(x=\sqrt { 2 } \) we get
\(\left( \sqrt { 2+1 } \right) ^{ 6 }+\left( \sqrt { 2-1 } \right) ^{ 6 }=2[\left( \sqrt { 2 } \right) ^{ 6 }+15\left( \sqrt { 2 } \right) ^{ 4 }+15\left( \sqrt { 2 } \right) ^{ 2 }+1\)
\(=2\left( { 2 }^{ 3 }+15\times { 2 }^{ 2 }+15\times 2+1 \right) =2(8+15\times 4+30+1)\)
\(=2(8+60+30+1)=2\times 99=198\)

\(T_{ r+1 }={ ^{ 7 }C }_{ r }\quad 3^{ 7-r }\times \quad x^{ 7-2r }\quad (-1)^{ r }\)
For coefficient of x^3, put 7-2r=3 \(\Rightarrow \)  r=2
\({ T }_{ 3 }=T_{ 2+1 }={ ^{ 7 }C }_{ r }\quad 3^{ 7-2 }\quad x^{ 7-4 } =5103\)

The general term in the expansion of (1-y)^-n is
\({ T }_{ r+1 }=\frac { n(n+1)(n+2)...(n+r-1) }{ r! } .{ y }^{ r }\)
\(\therefore \) In the expansion of(1-2x)^-5/2, we have
\({ T }_{ r+1 }=\frac { \left( \frac { 5 }{ 2 } \right) \left( \frac { 5 }{ 2 } +1 \right) \left( \frac { 5 }{ 2 } +2 \right) ...\left( \frac { 5 }{ 2 } +r-1 \right) }{ r! } .\left( 2x \right) ^{ r }\)
\(=\frac { 5.7.9...(3+2r) }{ { 2 }^{ r }(r!) } .{ 2 }^{ r }.{ x }^{ r }\)
\(Putting\quad r=6,\quad we\quad get\)
\({ T }_{ 7 }=\frac { 5.7.9.11.13.15 }{ { 2 }^{ 6 }.\left( 6! \right) } .{ 2 }^{ 6 }.{ x }^{ 6 }=\frac { 15015 }{ 16 } ={ x }^{ 6 }\)
Hence, coefficient of x⁶expansion of (1-2x)^-5/x is \(\frac { 15015 }{ 16 } \).

The general term in the expansion of \(\left( 2x+\frac { y }{ 3 } \right) ^{ 15 }\) is
\({ T }_{ r+1 }=^{ 15 }{ C }_{ r }(2x)^{ 15-r }\left( \frac { y }{ 3 } \right) ^{ r }[\because { T }_{ r+1 }=^{ n }{ C }_{ r }{ a }^{ n-r }{ b }^{ r }]\)
\(=^{ 15 }{ C }_{ r }{ 2 }^{ 15-r }\times { 3 }^{ -r }\times { x }^{ 15-r }{ y }^{ r }\)
For determining 7th term, put r=6, we get
\({ T }_{ 6+1 }=^{ 15 }{ C }_{ 6 }2^{ 15-6 }\times { 3 }^{ -6 }\times { x }^{ 15-6 }{ y }^{ 6 }\)
\(=^{ 15 }{ C }_{ 6 }{ 2 }^{ 9 }\times { 3 }^{ -6 }\times { x }^{ 9 }{ y }^{ 6 }\)

The general term in the expansion of (3x²+4y)¹⁰ is 
\({ T }_{ r+1 }=^{ 10 }{ C }_{ r }(3{ x }^{ 2 })^{ 10-r }(4y)^{ r }[\because { T }_{ r+1 }=^{ n }{ C }_{ r }{ a }^{ n-r }{ b }^{ r }]\)
\(=^{ 10 }{ C }_{ r }\times { 3 }^{ 10-r }\times { 4 }^{ r }\times { x }^{ 20-2r }{ y }^{ r }\)

We have, \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}=\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}\)
\(={ }^{4} C_{0}^{\circ}\left(1+\frac{x}{2}\right)^{4}-{ }^{4} C_{1}\left(1+\frac{x}{2}\right)^{3}\left(\frac{2}{x}\right)+{ }^{4} C_{2}\left(1+\frac{x}{2}\right)^{2} \)
\(\left(\frac{2}{x}\right)^{2}-{ }^{4} C_{3}\left(1+\frac{x}{2}\right)\left(\frac{2}{x}\right)^{3}+{ }^{4} C_{4}\left(\frac{2}{x}\right)^{4} \)
\(=\left(1+\frac{x}{2}\right)^{4}-4\left(1+\frac{x}{2}\right)^{3} \frac{2}{x}+\frac{4 \times 3}{2}\left(1+\frac{x}{2}\right)^{2} \frac{4}{x^{2}} \)
\(-4\left(1+\frac{x}{2}\right) \times \frac{8}{x^{3}}+\frac{16}{x^{4}} \)
\(=\left(1+\frac{x}{2}\right)^{4}-\frac{8}{x}\left(1+\frac{x}{2}\right)^{3}+\frac{24}{x^{2}}\left(1+\frac{x}{2}\right)^{2}\)
\(-\frac{32}{x^{3}}\left(1+\frac{x}{2}\right)+\frac{16}{x^{4}}\)
Now, on expending \(\left(1+\frac{x}{2}\right)^{4},\left(1+\frac{x}{2}\right)^{3} \cdot\left(1+\frac{x}{2}\right)^{2}\) 
We get, \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}=\left(1+4 \cdot \frac{x}{2}+6 \cdot \frac{x^{2}}{4}+4 \cdot \frac{x^{3}}{8}+\frac{x^{4}}{16}\right) \)
\(-8 \cdot \frac{1}{x}\left(1+3 \cdot \frac{x}{2}+3 \cdot \frac{x^{2}}{4}+\frac{x^{3}}{8}\right) \)
\(+24 \cdot \frac{1}{x^{2}}\left(1+x+\frac{x^{2}}{4}\right)-32 \times \frac{1}{x^{3}}\left(1+\frac{x}{2}\right)+\frac{16}{x^{4}} \)
\(=\left(1+2 x+\frac{3 x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}\right)-\left(\frac{8}{x}+12+6 x+x^{2}\right) \)
\(+\left(\frac{24}{x^{2}}+\frac{24}{x}+6\right)-\left(\frac{32}{x^{3}}+\frac{16}{x^{2}}\right)+\frac{16}{x^{4}} \)
\(=\frac{x^{4}}{16}+\frac{x^{3}}{2}+x^{2}\left(\frac{3}{2}-1\right)+x(2-6)+(1-12+6) \)
\(+(24-8) \frac{1}{x}+(24-16) \frac{1}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}} \)
\(=\frac{x^{4}}{16}+\frac{x^{3}}{2}+\frac{x^{2}}{2}-4 x-5+\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}\)

Splitting 1.01 and using binomial theorem to write the first few terms we have
(1.01)^1000000 = (1 + 0.01)^1000000
= ^1000000C₀ + ^1000000C₁(0.01) + other positive terms
= 1 + 1000000 x 0.01 + other positive terms
= 1 + 10000 + other positive terms
> 10000
Hence (1.01)^1000000 > 10000.

(127)^1/3 = (125+2)^1/3
\(=\left( 125 \right) ^{ 1/3 }\left( 1+\frac { 2 }{ 125 } \right) ^{ 1/3 }\)
\(=5\left[ 1+\frac { 1 }{ 3 } \left( \frac { 2 }{ 125 } \right) +\frac { \frac { 1 }{ 3 } \left( \frac { 1 }{ 3 } -1 \right) }{ 2 } \left( \frac { 2 }{ 125 } \right) ^{ 2 }+... \right] \) 
= 5(1+0.0053+...) = 5.0265

(10.3)⁵ + (9.7)⁵ = (10+0.3)⁵ + (10-0.3)⁵
(10+0.3)⁵ = ⁵C₀ 10⁵ + ⁵C₁(10)⁴(0.3)¹ + ⁵C₂(10)³(0.3)² +......(i)
(10-0.3)⁵ =  ⁵C₀ 10⁵ - ⁵C₁(10)⁴(0.3)¹ + ⁵C₂(10)³(0.3)²....(ii)
On Edding equ (i) and (ii)
(10+0.3)⁵ + (10-0.3)⁵ = 2[⁵C₀ 10⁵+ ⁵C₂(10)³(0.3)² + ⁵C₄(10)¹(0.3)⁴]
Ans. 201800.81

The (r + 1)^th term of the expansion (x + y)ⁿ is given by \(\mathrm{T}_{r+1}={ }^{n} \mathrm{C}_{r} x^{n-r} y^{r}\)
For the 17^th term, we have, r + 1 = 17, i.e., r = 16
Therefore, \(\mathrm{T}_{17} =\mathrm{T}_{16+1}={ }^{50} \mathrm{C}_{16}(2)^{50-16} a^{16} \)
\(={ }^{50} \mathrm{C}_{16} 2^{34} a^{16 .} \)
Similarly, \(\mathrm{T}_{18} ={ }^{50} \mathrm{C}_{17} 2^{33} a^{17} \)
Given that \(\mathrm{~T}_{17} =\mathrm{T}_{18}\)
So \({ }^{50} \mathrm{C}_{16}(2)^{34} a^{16}={ }^{50} \mathrm{C}_{17}(2)^{33} a^{17}\)
Therefore \(\frac{{ }^{50} \mathrm{C}_{16} \cdot 2^{34}}{{ }^{50} \mathrm{C}_{17} \cdot 2^{33}}=\frac{a^{17}}{a^{16}}\)
i.e., \(a=\frac{{ }^{50} \mathrm{C}_{16} \times 2}{{ }^{50} \mathrm{C}_{17}}=\frac{50 !}{16 ! 34 !} \times \frac{17 ! \cdot 33 !}{50 !} \times 2=1\)

(1 + x)³ (1 + x)⁶
= (1 + x³ + 3x + 3x²) (1 + ⁶C₁ x + ⁶C₂ x² - ⁶C₃​​​ ​​​​x³ + ⁶C₄​​​ ​​​​x⁴ - ⁶C₅​​​ ​​​​x⁵ + ⁶C6​​​ ​​​​x⁶)
Coefficient of x⁵ = - ⁶C₅​​​ + ⁶C₂  + 3 ⁶C₄​​​ - 3 ⁶C₃
= - 6 + 15 + 45 - 60 = - 6

Here n = 9 which is odd
So the middle terms are \(\left( \frac { 9+1 }{ 2 } \right) and\left( \frac { 79+1 }{ 2 } +1 \right) \) th i.e. 5^th and 6^th terms.
The general term in the expansion of \(\left( 2x-\frac { { x }^{ 2 } }{ 6 } \right) \)is
\({ T }_{ r+1 }=^{ 9 }{ C }_{ r }(2x)^{ a-r }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }\)
Putting r = 4 and 5 in (i)
\({ T }_{ 5 }=^{ 9 }{ C }_{ 4 }(2x)^{ 9-4 }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }=^{ 9 }{ C }_{ 4 }(2x)^{ 5 }(-1)^{ 4 }\left( \frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }\)
\(=\frac { 9! }{ 4!5! } \times 32{ x }^{ 5 }\times \frac { { x }^{ 8 } }{ 1296 } =\frac { 9\times 8\times 7\times 6 }{ 4\times 3\times 2\times 1 } \times 32{ x }^{ 5 }\times \frac { { x }^{ 8 } }{ 1296 } =\frac { 28 }{ 9 } x^{ 13 }\)
\({ T }_{ 6 }=^{ 9 }{ C }_{ 5 }(2x)^{ 9-5 }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 5 }=^{ 9 }{ C }_{ 5 }(2x)^{ 4 }(-1)^{ 5 }\left( \frac { { x }^{ 2 } }{ 6 } \right) ^{ 5 }\)
\(=\frac { 9! }{ 5!4! } \times 16{ x }^{ 4 }\times \frac { { x }^{ 10 } }{ 7776 } =-\frac { 7 }{ 27 } { x }^{ 14 }\) 

We have \(\left[ 1+\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \right] ^{ 4 }\)
=\(\left[ 1+\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \right] ^{ 4 }\)
=⁴C₀+⁴C₁\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \)+⁴C₂\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^2\)+⁴C₃\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^3\)+⁴C₄\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) ^4\)
=1+4\(\left( \frac { x }{ 2 } -\frac { 2 }{ x } \right) \)+6\(\left( \frac { x^{ 2 } }{ 2 } -\frac { 2 }{ { x }^{ 2 } } -2 \right) \)+4\(\left( \frac { { x }^{ 3 } }{ 8 } -\frac { 8 }{ { x }^{ 3 } } -\frac { 3x }{ 2 } +\frac { 6 }{ x } \right) +\left[ ^{ 4 }C_{ 0 }\left( \frac { x }{ 2 } \right) ^{ 4 }-^{ 4 }C_{ 1 }\left( \frac { x }{ 2 } \right) ^{ 3 }\left( \frac { 2 }{ x } \right) +^{ 4 }C_{ 2 }\left( \frac { x }{ 2 } \right) ^{ 2 }\left( \frac { 2 }{ x } \right) ^{ 2 }-^{ 4 }C_{ 3 }\left( \frac { x }{ 2 } \right) \left( \frac { 2 }{ x } \right) ^{ 3 }+^{ 4 }C_{ 4 }\left( \frac { x }{ 2 } \right) ^{ 4 } \right] \)
=1+\(\left( 2x-\frac { 8 }{ x } \right) +\left( \frac { 3 }{ 2 } { x }^{ 2 }+\frac { 24 }{ { x }^{ 2 } } -12 \right) +\left( \frac { { x }^{ 3 } }{ 2 } -\frac { 32 }{ { x }^{ 3 } } -6x+\frac { 24 }{ x } \right) +\left( \frac { { x }^{ 4 } }{ 16 } -{ x }^{ 2 }+6\frac { 16 }{ { x }^{ 2 } } +\frac { 16 }{ { x }^{ 4 } } \right) \)
=-5-4x + \(\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 2 } +\frac { { x }^{ 4 } }{ 16 } +\frac { 16 }{ x } +\frac { 8 }{ { x }^{ 2 } } +\frac { 32 }{ { x }^{ 3 } } -\frac { 16 }{ { 4 }^{ 4 } } \)

In the expansion of (1+x)²ⁿ, the general term is
T_r+1 =²ⁿC_r x^r
For coefficient of xⁿ , put r=n, we get T_n+1 =²ⁿC_n xⁿ
\(\therefore \)  Coefficient of xⁿ in the expansion of (1+x)²ⁿ =T_r+1 =²ⁿC_n  
Now, in the expansion of (1+x)^2n-1, the general term is
T_r+1 =^2n-1C_r  x^r
\(\therefore \) For coefficient of xⁿ in the expansion of (1+x)^2n-1
=^2n-1C_n

As 2n is even, the middle term in the expansion of (1+x)²ⁿ is given by \(\left( \frac { 2n }{ 2 } +1 \right) \)th term i.e. (n+1)th term,
Now,\(T_{ n+1 }={ ^{ 2n }C }_{ n }(1)^{ 2n-n }(x)^{ n }={ ^{ 2n }C }_{ n }x^{ n }=\frac { (2n)! }{ n!(2n-n)! } .x^{ n }\)
\(=\frac { (2n)! }{ n!n! } .x^{ n }=\frac { 2n(2n-1)(2n-2)...4.3.2.1 }{ n!n! } .x^{ n }\)
\(=\frac { 1.2.3.4...(2n-2)(2n-1)(2n) }{ n!n! } .x^{ n }\)
\(=\frac { \left[ 1.3.5...(2n-1) \right] \left[ 2.4.6...(2n) \right] }{ n!n! } .x^{ n }\)
\(=\frac { \left[ 1.3.5...(2n-1) \right] .2^{ n } }{ n! } .x^{ n }\) 

Suppose three consecutive terms in the expansion of (1+a)ⁿ are (r-1)th, rth and (r+1)th terms.
Then, \(T_{ r-1 }={ ^{ n }C }_{ r }-2^{ a^{ r-2 } }\quad \Rightarrow T_{ 2 }={ ^{ n }C }_{ r-1 }\quad a^{ r-1 }\)
and  \(T_{ r+1 }={ ^{ n }C }_{ r }a^{ r }\)
Since, coefficients are in the ratio 1:7:42, so we have
\(\frac { Coefficient\quad of\quad Tr-1 }{ Cofficient\quad of\quad { T }_{ r } } =\frac { 1 }{ 7 } \Rightarrow \frac { { ^{ n }C }_{ r-2 } }{ { ^{ n }C }_{ r-1 } } =\frac { 1 }{ 7 }\)
\(\Rightarrow \frac { \frac { n! }{ (r-2)!(n-r+2)! } }{ \frac { n! }{ (r-1)!(n-r+1)! } } =\frac { 1 }{ 7 } \Rightarrow 8r-n=9\quad ....(1)\)
Similarly,  \(\frac { Coefficient\quad of\quad T_{ r } }{ Coefficient\quad of\quad { T }_{ r+1 } } =\frac { 7 }{ 42 } \Rightarrow \frac { { ^{ n }C }_{ r-1 } }{ { ^{ n }C }_{ r } } =\frac { 7 }{ 42 } \)
\(\Rightarrow \frac { \frac { n! }{ (r-1)!(n-r+1)! } }{ \frac { n! }{ r!(n-r)! } } =\frac { 1 }{ 6 } \Rightarrow 7r-n=1\quad ....(ii)\)
solve Eq. (i) and (ii) 
Ans. n=35

(1.2)⁴⁰⁰⁰ = (1 +0.2)⁴⁰⁰⁰
= ⁴⁰⁰⁰C₀ +   ⁴⁰⁰⁰C₁(0.2) + sum of positive terms
= 1 + 4000(0.2) + a positive number
= 1 + 800 + positive number
= 800.

\({ \left( 2-3x \right) }^{ -3 }={ 2 }^{ -3 }{ \left( 1-\frac { 3 }{ 2 } \right) }^{ -3 }=\frac { 1 }{ 8 } { \left( 1-\frac { 3 }{ 2 } \right) }^{ -3 }\) 
Now expand by
\({ \left( 1-x \right) }^{ -n }=1+nx+\frac { n\left( n+1 \right) }{ 2! } { x }^{ 2 }+\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3! } { x }^{ 3 }+......\)