Here, a=2x, b=3y and n = 5
Given,\((2x+3y)^{ 5 }\)
\(=^{ 5 }{ C }_{ o }\left( 2x \right) ^{ 5 }+^{ 5 }{ C }_{ 1 }\left( 2x \right) ^{ 4 }(3y)^{ 1 }+^{ 5 }{ C }_{ 2 }\left( 2x \right) ^{ 3 }(3y)^{ 2 }+^{ 5 }{ C }_{ 3 }\left( 2x \right) ^{ 2 }\times (3y)^{ 3 }+^{ 5 }{ C }_{ 4 }\times \left( 2x \right) ^{ 1 }\times (3y)^{ 4 }+^{ 5 }{ C }_{ 5 }\times \left( 2x \right) ^{ o }\times (3y)^{ 5 }\)
\(=32{ x }^{ 5 }+240{ x }^{ 4 }y+720{ x }^{ 3 }{ y }^{ 2 }+1080{ x }^{ 2 }{ y }^{ 3 }+810x{ y }^{ 4 }+243{ y }^{ 5 }\)

\(\left( 9-6x \right) ^{ -\frac { 3 }{ 2 } }=(9)^{ -\frac { 3 }{ 2 } }(1-\frac { 2 }{ 3 } x)^{ -\frac { 3 }{ 2 } }\)
\(=\left( \frac { 1 }{ 27 } +\frac { 1 }{ 27 } x+\frac { 5 }{ 162 } { x }^{ 2 }+\frac { 35 }{ 1458 } { x }^{ 3 }+... \right) ,\quad when\quad \left| x \right| <\frac { 3 }{ 2 } \)

(1+x)ⁿ (1-x)ⁿ =(1-x²)ⁿ
Ans: T₃ =ⁿC₂

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
\(\left( \sqrt { 2+1 } \right) ^{ 6 }+\left( \sqrt { 2-1 } \right) ^{ 6 }=2[\left( \sqrt { 2 } \right) ^{ 6 }+15\left( \sqrt { 2 } \right) ^{ 4 }+15\left( \sqrt { 2 } \right) ^{ 2 }+1\)
\(=2\left( { 2 }^{ 3 }+15\times { 2 }^{ 2 }+15\times 2+1 \right) =2(8+15\times 4+30+1)\)
\(=2(8+60+30+1)=2\times 99=198\)

We have, \(\left( 102 \right) ^{ 5 }=\left( 100+2 \right) ^{ 5 }\)
\(=^{ 5 }{ C }_{ 0 }\left( 100 \right) ^{ 5 }+^{ 5 }{ C }_{ 1 }\left( 100 \right) ^{ 4 }\left( 2 \right) ^{ 1 }+^{ 5 }{ C }_{ 2 }\left( 100 \right) ^{ 3 }\left( 2 \right) ^{ 2 }+^{ 5 }{ C }_{ 3 }\left( 100 \right) ^{ 2 }\left( 2 \right) ^{ 3 }+^{ 5 }{ C }_{ 4 }\left( 100 \right) \left( 2 \right) ^{ 4 }+^{ 5 }{ C }_{ 5 }\left( 2 \right) ^{ 5 }\)
\(=^{ 5 }{ C }_{ 0 }\left( 10 \right) ^{ 10 }+^{ 5 }{ C }_{ 1 }\left( 10 \right) ^{ 8 }2+^{ 5 }{ C }_{ 2 }\left( 10 \right) ^{ 6 }\times 4+^{ 5 }{ C }_{ 2 }\left( 10 \right) ^{ 4 }\times 8+^{ 5 }{ C }_{ 1 }\left( 10 \right) ^{ 2 }\times 16+^{ 5 }{ C }_{ 2 }\times 32\)
\(=1\times { 10 }^{ 10 }+5\times { 10 }^{ 8 }\times 2+\frac { 5\times 4 }{ 2 } \times { 10 }^{ 6 }\times 4+\frac { 5\times 4 }{ 2 } \times { 10 }^{ 4 }\times 8+5\times 100\times 16+32\) 
= 10000000000+1000000000+40000000+800000+8000+32
= 11040808032

\(\left( y+b \right) ^{ n }=^{ n }{ { C }_{ 0 }{ y }^{ n } }+^{ n }{ { C }_{ 1 }{ y }^{ n-1 }b+^{ n }{ { C }_{ 2 }{ y }^{ n-2 }{ b }^{ 2 }+.... } }+^{ n }{ { C }_{ n }{ b }^{ n } }\)
\(=P+Q\) .....(i)
\( \left( y-b \right) ^{ n }=^{ n }{ { C }_{ 0 }{ y }^{ n } }-^{ n }{ { C }_{ 1 }{ y }^{ n-1 }b+^{ n }{ { C }_{ 2 }{ y }^{ n-2 }{ b }^{ 2 }-...... } }+^{ n }{ { C }_{ n }{ b }^{ n } }\)
\(=P+Q\) ...(ii)
Squaring Eqs. (i) and (ii) then adding, we get
\(\left( y+b \right) ^{ 2n }+\left( y-b \right) ^{ 2n }={ \left( P+Q \right) }^{ 2 }+{ \left( P-Q \right) }^{ 2 }\)
\(={ P }^{ 2 }+Q^{ 2 }+2PQ+{ P }^{ 2 }+Q^{ 2 }-2PQ=2\left( { P }^{ 2 }+Q^{ 2 } \right) \)

Here, n=10 (even) So there will be one one middle term i.e.\(\left( \frac { 10+2 }{ 2 } \right) \) th term or 6th term
T₆=T₅₊₁ =¹⁰C₅ \(\left( \frac { x }{ 3 } \right) ^{ 10-5 }\)(9y)⁵ =61236 x⁵y⁵

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
\({ T }_{ 6+1 }=^{ 15 }{ C }_{ 6 }2^{ 15-6 }\times { 3 }^{ -6 }\times { x }^{ 15-6 }{ y }^{ 6 }\)
\(=^{ 15 }{ C }_{ 6 }{ 2 }^{ 9 }\times { 3 }^{ -6 }\times { x }^{ 9 }{ y }^{ 6 }\)

Given expression is (x+3y)² . Here, n = 2
\(\therefore \) The number of terms in expansions is (n+1)
i.e 2+1 = 3

11ⁿ - 10ⁿ = (1+10)ⁿ - 10n
= 1 + 10n + ⁿC₂(10)² + ⁿC₃(10)³ + -10n
= 1 + 100 {ⁿC₂ + ⁿC₃ 10+...+ ⁿC_n10^n-2}
= 100 x an integer + 1
\(\Rightarrow \) 11ⁿ - 10n leaves remainder 1 when divided by 100.

(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

\(\left( c+dy \right) ^{ -2 }={ c }^{ -2 }(1+\frac { d }{ c } y)^{ -2 }\)
\(\therefore c^{ -2 }[1-2\times \frac { d }{ c } y+\frac { \left( -2 \right) \left( -2-1 \right) }{ 2! } \left( \frac { d }{ c } y \right) ^{ 2 }+...]=\frac { 1 }{ 4 } -3y+...\)
\(\Rightarrow { c }^{ -2 }-\frac { 2d }{ { c }^{ 3 } } y+...=\frac { 1 }{ 4 } -3y+...\)
\(On\quad comparing\quad first\quad term,{ c }^{ -2 }=\frac { 1 }{ 4 } \Rightarrow \frac { 1 }{ { c }^{ 2 } } =\frac { 1 }{ 4 } \Rightarrow c=2\)
\(On\quad comparing\quad second\quad term,\quad d=12\)
\( Ans.2,12\)

\(\left( 2+3y \right) ^{ -5 }=[3y(1+\frac { 2 }{ 3y } )]^{ -5 },\quad when\quad \left| \frac { 2 }{ 3y } \right| <1\)
\(=\frac { 1 }{ 243{ y }^{ 5 } } (1+\frac { 2 }{ 3y } )^{ -5 }\)
\(=\frac { 1 }{ 243 } [\frac { 1 }{ { y }^{ 5 } } -\frac { 10 }{ { 3y }^{ 6 } } +\frac { 20 }{ { 3y }^{ 7 } } -\frac { 280 }{ { 27 }y^{ 8 } } +...],\quad when\quad \left| y \right| >\frac { 2 }{ 3 } \)

Here, r>1, n>2
\(\therefore \)  T_3r = ²ⁿC_3r-1 x^3r-1; T_{r+2 =} ²ⁿC_r+1 x^r+1
Then, ²ⁿC_3r-1= ²ⁿC_r+1 
\(\Rightarrow \) 3r - 1 + r + 1 = 2n \(\Rightarrow \) n = 2r

Here n = 7, which is odd.
So the middle terms are \(\left( \frac { 7+1 }{ 2 } \right) th,\left( \frac { 7+1 }{ 2 } +1 \right) th\) are 4^th and 5^th terms.
The general term in the expansion of\(\left( 3-\frac { { x }^{ 3 } }{ 6 } \right) ^{ 7 }\) is
\({ T }_{ r+1 }=^{ 7 }{ C }_{ r }{ (3) }^{ 7-r }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ r }....(i)\)
Putting r = 3 and 4 in (i)
\(\therefore \quad { T }_{ 4 }=^{ 7 }C_{ 3 }{ (3) }^{ 7-3 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 3 }{ (3) }^{ 4 }.{ (-1) }^{ 3 }.\frac { { x }^{ 9 } }{ { (6) }^{ 3 } } \)
\(=35\times 81\times -\frac { { x }^{ 9 } }{ 216 } =-\frac { 105 }{ 8 } { x }^{ 9 }\)
Now \({ T }_{ 5 }=^{ 7 }C_{ 4 }{ (3) }^{ 7-4 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 4 }{ (3) }^{ 3 }.(-1)^{ 4 }\frac { { x }^{ 12 } }{ { (6) }^{ 4 } } \)
\(=35\times 27\times \frac { { x }^{ 12 } }{ 1296 } =\frac { 35 }{ 48 } { x }^{ 12 }\)