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 T_r+1 = ¹⁵C_r (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 T_r+1 = C_rx^r

        Hence coefficient of (2r + 4)^th term will be

        T_2r+4 = T_2r+3+1 = ¹⁸C_2r+3

and coefficient or (r - 2)^th term will be

        T_r-2 = T_r-3+1 = ¹⁸C_r-3.

        => ¹⁸C_2r+3 = ¹⁸C_r-3.

        => (2r + 3) + (r-3) = 18 (·.· ⁿC_r = ⁿC_K => 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 = ⁿC_r

                a₂ = ⁿC_r+1

                a₃ = ⁿC_r+2 and

                a4 = ⁿC_r+3

        Now  a₁/(a₁+a₂) = ⁿC_r/(ⁿC_r+ⁿC_r+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 T_r+1 be the greatest term, then T_r < T_r+1 > T_r+2

        Consider : T_r+1 > T_r

                => ²⁰C_r   (1/√3)^r > 20C_r-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 T_r+1 > T_r+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:

        Show that  ⁿC_k (sin kx) cos (n-k)x = 2^n-1 sin(nx)

Solution:

        We have ⁿC_k sin kx cos (n-k)x

                =1/2 ⁿC_k [sin (k x + nx - kx) + sin (kx - nx + kx)]

                =1/2 ⁿC_k sin n x + 1/2 ⁿC_k sin (2kx - nx)

                = 1/2 sin n x ⁿC_k 1/2 [ⁿC₀ sin (-nx) + ⁿC₁ sin (2-n)x +...

                ...+ ⁿC_n-1 sin (n-2)x + ⁿC_n sin nx]

= 2^n-1 sin nx + 0 (as terms in bracket, which are equidistant, from end and beginning will cancel each other). 

(Hence, proved)