In the expansion of (1+a)^m+n
Coefficient of a^m = ^m+nC_m = \(\frac{(m+n)!}{m!(m+n-m)!}=\frac{(m+n)!}{m!n!}\)   --- (1)
Coefficient of aⁿ = ^m+nC_n = \(\frac{(m+n)!}{n!(m+n-m)!}=\frac{(m+n)!}{n!m!}\)  --- (2)
(1) = (2) ⇒ Coefficient of a^m = Coefficient of aⁿ.

Given series is 3 - 6 + 9 - 12+ ...
= 3 (1) + 6 (- 1) + 9 (-1)² + 1² (- 1)³+ . . .
This is an arithmetic geometric (AG) series with correspondingA.P 3, 6, 9, 12 ... and G.P 1, -1, (-1)²,(-1)³.
\(\therefore\). n^th term of the given A. G. series is
= (n^th term of 3, 6, 9, ... ) (n^th term of 1, - 1, (-1)², ... )
= [3 + (n - 1)3] [1 (-1)^n-1]      [\(\because\) For AP, a = 3, d = 3 for GP = a = 1, r = -1]
= (3 + 3n - 3) (-1)^n-1
= 3n (-1)^n-l.

Let T_n be the n^th term of the sequence
Then, given T₁ = 3, T₂ = 3, T₃ = 6 and 
T_n = T_n-1 + T_n-2, n > 2.
T₃ = T₂ + T₁ = 3 + 3 = 6
T₄ = T₃ + T₂ = 6 + 3 = 9
T₅ = T₄ + T₃ = 9 + 6 = 15.

Given \({ \left( x-\frac { 1 }{ 2y } \right) }^{ 10 }\)
Here n = 10, x = x and \(a=\left( \frac { -1 }{ 2y } \right) \)
Middle term = \({ T }_{ \frac { 10+2 }{ 2 } }={ T }_{ 6 }\)
General term is \({ T }_{ r+1 }=nCr{ x }^{ n-r }{ a }^{ r }\)
Putting r = 5 we get,
\({ T }_{ 6 }=10{ C }_{ 5 }{ x }^{ 10-5 }{ \left[ -\frac { 1 }{ 2y } \right] }^{ 5 }=\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } .{ x }^{ 5 }\left( \frac { -1 }{ 32.{ y }^{ 5 } } \right) \)
\(=-225.{ x }^{ 5 }.\frac { 1 }{ 32{ y }^{ 5 } } { T }_{ 6 }=\frac { -63{ x }^{ 5 } }{ 8{ y }^{ 5 } } \)

We have eⁿ = \(1+\frac { n }{ 1! } +\frac { n^{ 2 } }{ 2! } +\frac { n^{ n } }{ n! } \)
⇒ \({ e }^{ n }>\frac { { n }^{ n } }{ n! } \)
⇒ \(n!>\frac { { n }^{ n } }{ { e }^{ n } } \)
⇒ \(n!>\left( \frac { n }{ e } \right) ^{ n }\). Hence proved