Can you provide me some questions of Binomial Theorem - Model Problems for preparation of IIT JEE (Indian Institutes of Technology Joint Entrance Examination)?

Some questions of Binomial Theorem - Model Problems for preparation of IIT JEE (Indian Institutes of Technology Joint Entrance Examination) are as follows:

Example 1:
Find the coefficient of the independent term of x in expansion of (3x - (2/x2))15.
Solution:
The general term of (3x - (2/x2))15 is written, as Tr+1 = 15Cr(3x)15-r (-2/x2)r. It is independent of x if,
15 - r - 2r = 0 => r = 5
.•. T6 = 15C5(3)10(-2)5 = - 16C5 310 25.

Example 2:
If the coefficient of (2r + 4)th and (r - 2)th terms in the expansion of (1+x)18 are equal then find the value of r.
Solution:
The general term of (1 + x)n is Tr+1 = Crxr
Hence coefficient of (2r + 4)th term will be
T2r+4 = T2r+3+1 = 18C2r+3
and coefficient or (r - 2)th term will be
Tr-2 = Tr-3+1 = 18Cr-3.
=> 18C2r+3 = 18Cr-3.
=> (2r + 3) + (r-3) = 18 (•.• nCr = nCK => r = k or r + k = n)
.•. r = 6
Example 3:
If a1, a2, a3 and a4 are the coefficients of any four consecutive terms in the expansion of (1+x)n then prove that:
a1/(a1+a2) + a2/(a3+a4) = 2a2/(a2+a3)
Solution:
As a1, a2, a3 and a4 are coefficients of consecutive terms, then
Let a1 = nCr
a2 = nCr+1
a3 = nCr+2 and
a4 = nCr+3
Now a1/(a1+a2) = nCr/(nCr+nCr+1) = 1/(1+((n-r)/(r+1))) = (r+1)/(n+1)
Similarly, a2/(a2+a3) = (r+3)/(n+1)
Now a3/(a3+a4) + a1/(a1+a2) = (2r+4)/(n+1)
= 2(r+1)/(n+1) = 2a2/(a2+a3) (Hence, proved)
Example 4:
Find out which one is larger 9950 + 10050 or 10150.
Solution:
Let's try to find out 10150 - 9950 in terms of remaining term i.e.
10150 - 9950 = (100+1)50 - (100 - 1)50
= (C0.10050 + C110049 + C2.10048 +......)
= (C010050 - C110049 + C210048 -......)
= 2[C1.10049 + C310047 +.........]
= 2[50.10049 + C310047 +.........]
= 10050 + 2[C310047 +............]
> 10050
=> 10150 > 9950 + 10050
Example 5:
Find the value of the greatest term in the expansion of √3(1+(1/√3))20.
Solution:
Let Tr+1 be the greatest term, then Tr < Tr+1 > Tr+2
Consider : Tr+1 > Tr
=> 20Cr (1/√3)r > 20Cr-1(1/√3)r-1
=> ((20)!/(20-r)!r!) (1/(√3)r) > ((20)!/(21-r)!(r-1)!) (1/(√3)r-1)
=> r < 21/(√3+1)
=> r < 7.686 ......... (i)
Similarly, considering Tr+1 > Tr+2
=> r > 6.69 .......... (ii)
From (i) and (ii), we get
r = 7
Hence greatest term = T8 = 25840/9
Example 6:
Find the coefficient of x50 in the expansion of (1+x)1000 + 2x(1+x)999 + 3x2 (1+x)998 +...+ 1001x1000.
Solution:
Let S = (1 + x)1000 + 2x(1+x)999 +...+ 1000x999 (1+x) + 1001 x1000
This is an Arithmetic Geometric Series with r = x/(1+x) and d = 1.
Now (x/(1+x)) S = x(1 + x)999 + 2x2 (1 + x)998 +...+ 1000x1000 + 1000x1001/(1+x)
Subtracting we get,
(1 - (x/(x+1))) S =(1+x)1000 + x(1+x)999 +...+ x1000 - 1001x1000/(1+x)
or S = (1+x)1001 + x(1+x)1000 + x2(1+x)999 +...+ x1000 (1+x)-1001x1001
This is G.P. and sum is
S = (1+x)1002 - x1002 - 1002x1001
So the coeff. of x50 is = 1002C50

Example 7:
If (15+6√6)2n+1 = P, then prove that P(1 - F) = 92n+1 (where F is the fractional part of P).
Solution:
We can write
P = (15+6√6)2n+1 = I + F (Where I is integral and F is the fractional part of P)
Let F' = (15+6√6)2n+1

Note: 6√6 = 14.69

=> 0 < 156√6 < 1

=> 0 < (15+6√6)2n+1 < 1

=> 0 < F' < 1
Now, I + F = C0 (15)2n+1 + C1(15)2n 6√6 + C2 (15)2n (6√6 )2 +...
F' = C0 (15)2n+1 - C1(15)2n 6√6 + C2(15)2n-1 (6√6 )2 +...
I + F + F' = 2[C0 (15)2n+1 + C2 (15)2n-1 (6√6 )2 +...]
Term on R.H.S. is an even integer.
=> I + F + F' = Even integer
=> F + F' = Integer
But, 0 < F < 1 and (F is fraction part)
0 < F' < 1
=> 0 < F + F' < 2
Hence F + F' = 1
F' = (1-f)
.•. P(1-F) = (15 + 6√6 )2n+1 (15-6√6 )2n+1
= (9)2n+1 (Hence, proved)