Comparing \(\left( 2x+\frac { 3 }{ { x }^{ 2 } } \right) ^{ 9 }\)with (a+ b)ⁿ,
we have
a = 2x, b = \(\frac { 3 }{ { x }^{ 2 } } \) and n = 9
We know that
T_{r+ 1} = ⁿC_ra^n-r b^r
∴ T_r+1 = ⁹C_r (2x)^9-r\((\frac { 3 }{ { x }^{ 2 } } )^r\)
= ⁹C_r(2)^9-r (3)^rx^9-3r
Now 9 - 3r = 0 ⇒ 3r = 9 ⇒ r = 3
Thus T_{3+ 1} i.e. T₄ is the term independent of x, which is equal to ⁹C₃(2)⁶(3)³=19736

n = 7, a = 2, x = 1

7, 14

2(5a⁴b + 10a²b³ + b⁵); 568\(\sqrt3\)

Comparing \(\left( x-\frac { { 1 } }{ 6^2 } \right) ^{ 24 }\)with(a+b)ⁿ,
we have 
a = x, b =\(\left( -\frac { 1 }{ { x }^{ 2 } } \right) \)and n=24
We know that 
T_r+1=ⁿC_ra^n-rb^r
\(\therefore \) T_r+1=²⁴C_rx^{24-r .\(\left( -\frac { 1 }{ { x }^{ 2 } } \right) ^r\)}
Now 24 -3r=6
\(\Rightarrow \)3r = 24 - 6 \(\Rightarrow \)  r = 6
\(\therefore \)  Coefficient of x⁶ = ²⁴C₆(- 1)⁶
= ²⁴C₆=​​​​​​​\(\frac { 24! }{ 18!6! } \)​​​​​​​

10

55

\((1-3x+7x^{ 2 })(1-x)^{ 16 }\)
\(=(1-3x+7x^{ 2 })({ ^{ 16 }C }_{ 0 }-{ ^{ 16 }C }_{ 1 }x+{ ^{ 16 }C }_{ 2 }x^{ 2 }+...)\)
\(=({ ^{ 16 }C }_{ 0 }-{ ^{ 16 }C }_{ 1 }x+{ ^{ 16 }C }_{ 2 }x^{ 2 }\quad -....)-3x({ ^{ 16 }C }_{ 0 }-{ ^{ 16 }C }_{ 1 }x+...)+7x^{ 2 }({ ^{ 16 }C }_{ 0 }-{ ^{ 16 }C }_{ 1 }x+...)\)
Here, the term containing x is
\({ ^{ 16 }C }_{ 1 }x-3{ ^{ 16 }C }_{ 0 }x=-16x-3x\)
\(\therefore \) Coefficient of x=-16-3=-19
Ans: -19

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

We have, \(\left( \frac { 2x }{ 3 } -\frac { 3 }{ 2x } \right) ^{ 4 }=^{ 4 }{ C }_{ o }\left( \frac { 2x }{ 3 } \right) ^{ 4 }-^{ 4 }{ C }_{ 1 }\left( \frac { 2x }{ 3 } \right) ^{ 3 }\left( \frac { 3 }{ 2x } \right) +^{ 4 }{ C }_{ 2 }\left( \frac { 2x }{ 3 } \right) ^{ 2 }\left( \frac { 3 }{ 2x } \right) ^{ 2 }-^{ 4 }{ C }_{ 3 }\left( \frac { 2x }{ 3 } \right) \left( \frac { 3 }{ 2x } \right) ^{ 3 }+^{ 4 }{ C }_{ 4 }\left( \frac { 3 }{ 2x } \right) ^{ 4 }\)
\(=1\times \frac { { 16x }^{ 4 } }{ 81 } -4\times \frac { { 8x }^{ 3 } }{ 27 } \left( \frac { 3 }{ 2x } \right) +6.\frac { { 4x }^{ 2 } }{ 9 } \left( \frac { 9 }{ { 4x }^{ 2 } } \right) -4\left( \frac { 2x }{ 3 } \right) \left( \frac { 27 }{ { 8x }^{ 3 } } \right) +1\times \left( \frac { 8 }{ 16x^{ 4 } } \right) \)
\(\left[ \because \ ^{ 4 }{ C }_{ o }=^{ 4 }{ C }_{ 4 }=1,^{ 4 }{ C }_{ 3 }=^{ 4 }{ C }_{ 1 }=4\quad and\quad ^{ 4 }{ C }_{ 2 }=\frac { 4! }{ 2!2! } =\frac { 4\times 3\times 2! }{ 2\times 1\times 2! } =6 \right] \)
\(\frac { 16 }{ 81 } { x }^{ 4 }-\frac { 16 }{ 9 } { x }^{ 2 }+6-\frac { 9 }{ { x }^{ 2 } } +\frac { 81 }{ { 16x }^{ 4 } } \)

\({ \left( y+\frac { 1 }{ y } \right) }^{ 6 }-{ \left( y-\frac { 1 }{ y } \right) }^{ 6 }=2\)
\( \left[ ^{ 6 }{ { C }_{ 1 }{ y }^{ 6-1 } }{ \left( \frac { 1 }{ y } \right) }^{ 1 }+^{ 6 }{ { C }_{ 3 }{ y }^{ 6-3 } }{ \left( \frac { 1 }{ y } \right) }^{ 3 }+^{ 6 }{ { C }_{ 5 }{ y }^{ 6-5 } }{ \left( \frac { 1 }{ y } \right) }^{ 5 } \right] \)
\(=2\left[ 6{ y }^{ 5 }.\frac { 1 }{ y } +20{ y }^{ 3 }.\frac { 1 }{ { y }^{ 3 } } +6y.\frac { 1 }{ { y }^{ 5 } } \right] =12\left( { y }^{ 4 }+\frac { 1 }{ { y }^{ 4 } } +\frac { 10 }{ 3 } \right) \)

\({ \left( a+b \right) }^{ 4 }-{ \left( a-b \right) }^{ 4 }=2\left[ ^{ 4 }{ { C }_{ 1 }{ a }^{ 3 }b^{ 1 }+^{ 4 }{ { C }_{ 3 }{ ab }^{ 3 } } } \right] \)
\(=2\left[ { 4{ a }^{ 3 }b^{ 1 }+4{ { ab }^{ 3 } } } \right] \)
\(=8\left( { a }^{ 3 }b+{ { ab }^{ 3 } } \right) \) 
Put \(a=\sqrt { 3 } ,b=\sqrt { 2 }\) to get the required answer.

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 (1 + x)²ⁿ
∴T_m+1=²ⁿC_m(x)^m
∴ Coefficient of (m + 1)^th term = ²ⁿC_m
Also T_m+3 = ²ⁿC_m+2 (x)^{m + 2}
∴ Coefficient of (m + 3)^th term = ²ⁿC_{m + 2}
It is given that
²ⁿC_m =²ⁿC_m+2
∴ m + (m + 2) = 2n
⇒ 2m + 2 = 2n
⇒ m+ 1= n.

Here (x + c)ⁿ
=ⁿC₀ xⁿ+ⁿc₁ x^n-1a+ⁿC₂ x^n-2a² +... +ⁿC_naⁿ 
= P + Q....(i)
where P= ⁿC₀ xⁿ + ⁿC₂ x^n-2a² + ...
Q=ⁿC₁ x^{n - 1}a + ⁿC₃ x^{n- 3} a³ + ...
Also (x-a)^n
nC₀ xⁿ - ⁿC₁ x^{n - 1}a + ⁿC₂ x^n-2a² +...+(-1)^{n n}C_naⁿ 
=P - Q....(ii)
(i) Squaring and adding (i) and (ii), we have
(x + a)²ⁿ + (x - a)²ⁿ = (P + Q)² + (P - Q)²
= P² + Q² + 2PQ + p² +Q²-2PQ
= 2P²+ 2Q² =2 (P²+ Q²)
(ii)Multiplying (i) and (ii), we have
(x + a)ⁿ (x - a)ⁿ = (P + Q) (P - Q)
(x²-a²)ⁿ = P²- Q².

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)
\({ T }_{ 5 }=^{ 9 }{ C }_{ 4 }(2x)^{ 9-4 }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }=^{ 9 }{ C }_{ 4 }(2x)^{ 5 }(-1)^{ 4 }\left( \frac { { x }^{ 2 } }{ 6 } \right) ^{ 4 }\)
\(=\frac { 9! }{ 4!5! } \times 32{ x }^{ 5 }\times \frac { { x }^{ 8 } }{ 1296 } =\frac { 9\times 8\times 7\times 6 }{ 4\times 3\times 2\times 1 } \times 32{ x }^{ 5 }\times \frac { { x }^{ 8 } }{ 1296 } =\frac { 28 }{ 9 } x^{ 13 }\)
\({ T }_{ 6 }=^{ 9 }{ C }_{ 5 }(2x)^{ 9-5 }\left( -\frac { { x }^{ 2 } }{ 6 } \right) ^{ 5 }=^{ 9 }{ C }_{ 5 }(2x)^{ 4 }(-1)^{ 5 }\left( \frac { { x }^{ 2 } }{ 6 } \right) ^{ 5 }\)
\(=\frac { 9! }{ 5!4! } \times 16{ x }^{ 4 }\times \frac { { x }^{ 10 } }{ 7776 } =-\frac { 7 }{ 27 } { x }^{ 14 }\)