Here a₁ = 4, and a_n+1 = 2na_n.
Putting n = 1, a₂ = 2 \(\times\) 1\(\times\) a₁ = 2 \(\times\) 1\(\times\) 4 = 8
Putting n = 2, a₃ =2\(\times\)2\(\times\)a₂ = 4 \(\times\) 8 = 32
Putting n = 3, a₄ = 2\(\times\)3\(\times\)a₃ = 6 \(\times\) 32 = 192
Putting n = 4, a₅ = 2\(\times\)4\(\times\)a₄ = 8 \(\times\) 192 = 1536
Putting n = 5, a₆ = 2\(\times\) 5\(\times\) a₅ = 10\(\times\)1536 = 15360

Let the number be a and b
∴ a + b = \(\frac { 13 }{ 6 } \)
Let A₁, A₂,..A_2n be the 2n A,M s between a and b ....(1)
= \(2n\left( \frac { a+b }{ 2 } \right) =n(a+b)=n\times \frac { 13 }{ 6 } (2)\)   using (1)
Also  A₁+ A₂+ A_2n = 2n+1(given)   ...(3)
From (2)  and (3),\(\frac { 13n }{ 6 } \) = 2n+1
⇒   13n = 12n + 6
⇒    n = 6
∴  No of A.M's inserted = 2n - 2(6) = 12

 Given Series 1 + (a+bx) + \(\frac { (a+bx) }{ 2! } +\frac { (a+bx)^{ 3 } }{ 3! } \)
e^a+bx [using exponential series]
e^a . e^bx 
e^a  \(\left[ 1+\frac { bx }{ 1! } +\frac { { bx }^{ 2 } }{ 2! } +\frac { { bx }^{ 3 } }{ 3! } +..\frac { { bx }^{ n } }{ n! } +...\infty \right] \)
\({ e }^{ a }+\frac { { e }^{ a }.bx }{ 1! } +\frac { { e }^{ a }.{ b }^{ 2 }{ x }^{ 2 } }{ 2! } +\frac { { e }^{ q }.{ b }^{ 3 }{ x }^{ 3 } }{ 3! } +..+\frac { { e }^{ a }.{ b }^{ n }{ x }^{ n } }{ n! } +...\infty \)
Co - efficient of xⁿ   in the given series is \(\frac { { e }^{ a }.{ b }^{ n } }{ n! } \)

Given 3 + 10 + 17 +...
a₁ = 3, d₁ = 10 - 3 = 7
∴ T_n = a₁(n-1)d₁ = 3 + (n-1)7 = 7n-4  ...(1)
Also, given 63 + 65 + 67+..
a₂ = 63, d₂ = 65 - 63 = 2
∴ T_n = a₂+(n-1)d₂ = 63 + (n-1) 2 = 2n + 61   ..(2)
Let n^th term of given series be equal
⇒   7n - 4 =2n + 61
⇒ 5n = 65     [From (1)and (2)]
⇒ n = 13

In (1 + x)ⁿ, (r + 1)^th term from the beginning.
T_r+1 = nC_r 1^n-r. x^r = nC_rx^r   ....(1)
Its co-efficient is nC_r
In (1 + x)ⁿ, there are (n + 1)terms
So, the (r +1)^th term from the end will have (n + 1) - (r + 1) = n - r terms
∴ T_n-r+1 = nC_n-r 1^n-(n-r).x^n-r = nC_n-rx^n-r  ...(2)
Its Co-efficient is  nC_n-r
From (1) and (2), the Co-efficient of (r + 1)th term from the beginning and from the end are equal