Solved Examples

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)

 Example 8:

        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)

Example 9:

        Using ∫₀¹(tx+a-x)ⁿ dx,prove that ∫₀¹x^k (1-x)^(n-k) dx=[ ⁿC_k (n+1)]^(-1)

Solution:

        The given integral can easily be evaluated, as follows:

                I = ∫₀¹(tx+a-x)ⁿ dx

                  = ∫₀¹((t-1)x+1)ⁿ  dx

                   = [((t-1)x+1)⁽ⁿ⁺¹⁾/((n+1)(t-1))]₀¹

         =(t⁽ⁿ⁺¹⁾-1)/(n+1)(t-1)=1/(1+n)  [1 + t + t² +...+ t^k +...+ tⁿ]      ...... (i)

         Also, I = ∫₀¹(tx+(1-x))ⁿ dx

             =∫₀^{1 n}C_k (1-x)^(n-K) t^k x^k  dx                                   ...... (ii)

        Comparing the coefficient of t^k from (i) and (ii), we get,

                     ∫₀^{1  n}C_k. x^k (1-x)^n-k dx=1/(n+1)

                    => ∫₀¹x^k (1-x)^n-k dx = 1/( ⁿC_k (n+1))              

                          (Hence, proved)

Example 10:

Prove that (-3)^{r-1 3n}C_2r-1= where k = 3n/2  and n is an even positive integer.

Solution:

           Since n is even integer, let n = 2m,

           k = 3n/2 = 6m/2 =3m

           The summation becomes,

           S =

           Now, (1+x)^6m = ^6mC₀ + ^6mC₁ x + ^6mC₂ x² +...

                                        ...+ ^6mC_6m-1 x^6m-1 + ^6mC_6m x^6m          ...... (i)

                (1-x)^6m = ^6mC₀ + ^6mC₁ (-x) + ^6mC₂ (-x)² +...

                                ...+ ^6mC_6m-1 (-x)^6m-1 + ^6mC_6m (-x)^6m          ...... (ii)

                => (1+x)^6m - (1-x)^6m = 2[^6mC₁x + ^6mC₃x² +...+ ^6mC_6m-1x^6m-2]

                ((1+x)^6m - (1-x)^6m)/2x = ^6mC₁ + ^6mC₃  x² +......+ ^6mC_6m-1  x^6m - 2

          Let x² = y

          =>((1 + √y)6m-(1-√y)6m)/2√y=^6mC₁+^6mC₃y  + ^6mC₃ y+......^6mC_6m-1 y^3m-1

          With y = -3, RHS becomes = S.

            LHS =  

          

                       (Hence, proved)

Example 11:

        If k and n are two positive integers and

                S_k = 1_k + 2_k +.........+ n_k

        Then show that

                m+¹C₁S₁ + m+¹C₂S₂ +......+ m+¹C_mS_m = (1+n)^m+1 - (1+n)

Solution:

        We have,

                (1 + p)^m+1 = ^m+1C₀ + ^m+1C₁ p + ^m+1C₂ p² + ^m+1C_m+1 p^m+1

        Putting p = 1, 2, 3, ........., n

       =>2^m+1 - 1      = ^m+1C₀ + ^m+1C₁(1) + ^m+1C₂(1)² +...+ ^m+1C_m(1)^m

           3^m+1 - 2^m+1 = ^m+1C₀ + ^m+1C₁(2) + ^m+1C₂(2)² +...+ ^m+1C_m(2)^m

           4^m+1 - 3^m+1 = ^m+1C₀ + ^m+1C₁(3) + ^m+1C₂(3)² +...+ ^m+1C_m(3)^m

             

          (1+n)^m+1-n^m+1 = ^m+1C₀(n) + ^m+1C₂ Σn + ^m+1C₂ Σ(n)² +...+ ^m+1C_m ∑(n)^m.

           Adding all these terms, we get

             (1+n)^m+1 -1 = ^m+1C₀(n) + ^m+1C₁S₁ + ^m+1C₂S₂ +...+ ^m+1C_mS_m

       =>^m+1C₁S₁ + ^m+1C₂S₂ +...+ ^m+1C_mS_m = (1+n)^m+1 - (1+n)         

                 (Hence, proved)

Example 12:

        Given that S_n = 1 + q + q² +....+ qⁿ

        and P_n = 1 + ((q+1)/2) + ((q+1)/2)² + ....+((q+1)/2)ⁿ

        show that ⁿ⁺¹C₁ + ⁿ⁺¹C₂S₁ + ⁿ⁺¹C₂S₂ +...+ ⁿ⁺¹C_n+1 S_n = 2ⁿ p_n.

Solution:

        Both S_n and P_n are geometric series

              S_n = 1 + q + q² +......qⁿ = 1-q^n-1/1-q                                   ...... (i)

              P_n = 1 + ((q+1)/2) + ((q+1)/2)² +.....+((q+1)/2)ⁿ

             =(1-((q+1)/2)ⁿ⁺¹)/(1-(q+1)/2)=1/2ⁿ(2ⁿ⁺¹-(q+1)ⁿ⁺¹)/(1-q)         ...... (ii)

             ⁿ⁺¹C_r+1S_r = ⁿ⁺¹C_r+1 ((1-q^r+1)/(1-q)) =  ⁿ⁺¹C_r+1 ((1/(1-q)) - (q^r+1/(1-q)))

             =  (1/(1-q))ⁿ⁺¹C_r+1 -  (1/(1-q)) ⁿ⁺¹C_r+1 q^r+1

             =>ⁿ⁺¹C_r+1 S_r=(1/(1-q))ⁿ⁺¹C_r+1- (1/(1-q))ⁿ⁺¹C_r+1 q^r+1

             => ∑ⁿ⁺¹C_r+1S_r = 2ⁿp_n                              (Hence, proved)

Example 13:

        Show that if C_r is the coefficient of x^r in (1+x)ⁿ, then

        C_r³  = Coefficient of xn yn in the expression of ((1+x)(1+y)(x+y))n

Solution:

        (1+x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² +...+ ⁿC_n xⁿ                        ...... (1)

        (1+y)ⁿ = ⁿC₀ + ⁿC₁ y + ⁿC₂ y² +...+ ⁿC_n yⁿ                         ...... (2)

        (x+y)ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 y + ⁿC₂ x^n-2 +...+ ⁿC_n yⁿ          ....... (3)

        writing (1+y)ⁿ as (y+1)ⁿ and expanding it in decreasing powers of y.

                (y+1)ⁿ = ⁿC₀yⁿ + ⁿC₁y^n-1 + ⁿC₂y^n-2 +...+ ⁿC_n            ....... (4)

        multiplying (1), (3) and (4) we get,

                (1+x)ⁿ (y+1)ⁿ (x+y)ⁿ

                = (ⁿC₀ + ⁿC₁x + ⁿC₂x² +...+ ⁿC_nxⁿ) × (ⁿC₀yⁿ + ⁿC₁y^n-1 +...+ ⁿC_n)

                × (ⁿC₀xⁿ + ⁿC₁x^n-1y +...+ ⁿC_nyⁿ)

        Now, coefficient of xⁿyⁿ in RHS = ⁿC₀³ + ⁿC₁³ + ⁿC₂³ +... ⁿC_n³

                = coefficient of xⁿyⁿ in LHS

                = coefficient of xⁿyⁿ in {(1+x)(1+y)(x+y)}ⁿ              

       (Hence, proved)

Example 14:

        If (1+x)ⁿ = C₀ + C₁x + C₂x² +...+ C_nxⁿ (nεN)

        Then show that k³ [C_k/C_k-1] = 1/12 (n)(n+1)²(n+2)

Solution:

         C_r/C_r-1 = (n!/(n-r)!r!) (((r-1)!(n-r+1)!)/n!) = (((n+1)/r)-1)

        r³(C_r/C_r-1) = ((n+1)/r-1) r³ = (n+1)r² - r³

        =

        (n+1)n(n+1)(2n+1)/6-(n² (n+1)²)/4

        = n(n+1)/12 [4n² + 6n + 2 - 3n² - 3n] = n(n+1)(n²+3n+2)/12

        n(n+1)(n+1)(n+2)/12=(n(n+1)² (n+2))/12                                          

(Hence, proved)

Example 15:

        If (1+x)ⁿ = C₀ + C₁ x + C₂ x² +...+ C_n xⁿ then show that

        (i) ∑_0≤i<j≤n  C_i C_j = 2^2n-1 - 

        (ii) ∑_0≤i<j≤n  (C_i + C_j)² = (n - 1) ²ⁿC_n + 2²ⁿ

        (iii) ∑_0≤i<j≤n   ∑C_i  C_{j =} n (2^2n-1 - 1/2 ²ⁿC_n)

Solution:

(i)     We have

        (C₀ + C₁ + C₂ +...+ C_n)² = C₀² + C₁² +...+ C_n² + 2 ∑_0≤i<j≤n   ∑C_i  C_j

        we get, (2n)² = ²ⁿC_n + 2 ∑_0≤i<j≤n   ∑C_i  C_j

        therefore  ∑_0≤i<j≤n   ∑C_i  C_j = 2^2n-1 - (2n)!/2(n!)²                         

(Hence, proved)

(ii)   ∑_0≤i<j≤n   ∑C_i  C_j  n(C₀² + C₁² + C₂² +...+ Cⁿ²) + 2∑_0≤i<j≤n   ∑C_i  C_j

                                = n ²ⁿC_n + 2{2^2n-1 - (2n)!/2(n!)²}               [from part (i)]

                                = n ²ⁿC_n + 2²ⁿ - ²ⁿC_n

                                = (n-1) ²ⁿC_n + 2²ⁿ                                 

(Hence, proved)

(iii)    Let ∑_0≤i<j≤n   ∑C_i  C_j

        replace i by (n - i) and j by (n - j), we have

                P = ∑_0≤i<j≤n   ∑(2n-i0j) C_n-i  C_n-j

                =  ∑_0≤i<j≤n   ∑(2n-(i+j)) C_i C_j       [·.· ⁿC_n = ⁿC_n-r]

                = 2n ∑_0≤i<j≤n  ∑ C_i C_j - P

        .·.     ∑_0≤i<j≤n   ∑(i+j) C_i C_j = n[2^2n-1 - (2n)!/2(n!)²]                     

(Hence, proved)  

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