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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 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:

        Show that  symbol2 nCk (sin kx) cos (n-k)x = 2n-1 sin(nx)

Solution:

        We have symbol2nCk sin kx cos (n-k)x

                =1/2 symbol2 nCk [sin (k x + nx - kx) + sin (kx - nx + kx)]

                =1/2symbol2 nCk sin n x + 1/2 symbol2 nCk sin (2kx - nx)

                = 1/2 sin n x symbol2 nCk 1/2 [nC0 sin (-nx) + nC1 sin (2-n)x +...

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

= 2n-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 ∫01xk (1-x)(n-k) dx=[ nCk (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 +...+ tk +...+ tn]      ...... (i)

         Also, I = ∫01(tx+(1-x))n dx

             =∫01symbol2 nCk (1-x)(n-K) tk xk  dx                                   ...... (ii)

        Comparing the coefficient of tk from (i) and (ii), we get,

                     ∫01  nCk. xk (1-x)n-k dx=1/(n+1)

                    => ∫01xk (1-x)n-k dx = 1/( nCk (n+1))              

                          (Hence, proved)

Example 10:

Prove that equation17 (-3)r-1 3nC2r-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 = equation16

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

                                        ...+ 6mC6m-1 x6m-1 + 6mC6m x6m          ...... (i)

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

                                ...+ 6mC6m-1 (-x)6m-1 + 6mC6m (-x)6m          ...... (ii)

                => (1+x)6m - (1-x)6m = 2[6mC1x + 6mC3x2 +...+ 6mC6m-1x6m-2]

                ((1+x)6m - (1-x)6m)/2x = 6mC1 + 6mC3  x2 +......+ 6mC6m-1  x6m - 2

          Let x2 = y

          =>((1 + √y)6m-(1-√y)6m)/2√y=6mC1+6mC3y  + 6mC3 y+......6mC6m-1 y3m-1

          With y = -3, RHS becomes = S.

            LHS =  

           equation15

                       (Hence, proved)

Example 11:

        If k and n are two positive integers and

                Sk = 1k + 2k +.........+ nk

        Then show that

                m+1C1S1 + m+1C2S2 +......+ m+1CmSm = (1+n)m+1 - (1+n)

Solution:

        We have,

                (1 + p)m+1 = m+1C0 + m+1C1 p + m+1C2 p2 + m+1Cm+1 pm+1

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

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

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

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

              symbol4

          (1+n)m+1-nm+1 = m+1C0(n) + m+1C2 Σn + m+1C2 Σ(n)2 +...+ m+1Cm ∑(n)m.

           Adding all these terms, we get

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

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

                 (Hence, proved)

Example 12:

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

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

        show that n+1C1 + n+1C2S1 + n+1C2S2 +...+ n+1Cn+1 Sn = 2n pn.

Solution:

        Both Sn and Pn are geometric series

              Sn = 1 + q + q2 +......qn = 1-qn-1/1-q                                   ...... (i)

              Pn = 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+1Cr+1Sr = n+1Cr+1 ((1-qr+1)/(1-q)) =  n+1Cr+1 ((1/(1-q)) - (qr+1/(1-q)))

             =  (1/(1-q))n+1Cr+1 -  (1/(1-q)) n+1Cr+1 qr+1

             =>symbol3n+1Cr+1 Sr=(1/(1-q))symbol3n+1Cr+1- (1/(1-q))symbol3n+1Cr+1 qr+1

             => ∑n+1Cr+1Sr = 2npn                              (Hence, proved)

Example 13:

        Show that if Cr is the coefficient of xr in (1+x)n, then

        symbol3Cr3  = Coefficient of xn yn in the expression of ((1+x)(1+y)(x+y))n

Solution:

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

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

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

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

                (y+1)n = nC0yn + nC1yn-1 + nC2yn-2 +...+ nCn            ....... (4)

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

                (1+x)n (y+1)n (x+y)n

                = (nC0 + nC1x + nC2x2 +...+ nCnxn) × (nC0yn + nC1yn-1 +...+ nCn)

                × (nC0xn + nC1xn-1y +...+ nCnyn)

        Now, coefficient of xnyn in RHS = nC03 + nC13 + nC23 +... nCn3

                = 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 +...+ Cnxn (nεN)

        Then show that symbol2k3 [Ck/Ck-1] = 1/12 (n)(n+1)2(n+2)

Solution:

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

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

        = symbol5

        (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 +...+ Cn xn then show that

        (i) 0≤i<j≤n  Ci Cj = 22n-1symbol6

        (ii) 0≤i<j≤n  (Ci + Cj)2 = (n - 1) 2nCn + 22n

        (iii) 0≤i<j≤n   Ci  Cj = n (22n-1 - 1/2 2nCn)

Solution:

(i)     We have

        (C0 + C1 + C2 +...+ Cn)2 = C02 + C12 +...+ Cn2 + 2 0≤i<j≤n   Ci  Cj

        we get, (2n)2 = 2nCn + 2 0≤i<j≤n   Ci  Cj

        therefore  0≤i<j≤n   Ci  Cj = 22n-1 - (2n)!/2(n!)2                         

(Hence, proved)

(ii)   0≤i<j≤n   Ci  Cj  n(C02 + C12 + C22 +...+ Cn2) + 20≤i<j≤n   Ci  Cj

                                = n 2nCn + 2{22n-1 - (2n)!/2(n!)2}               [from part (i)]

                                = n 2nCn + 22n - 2nCn

                                = (n-1) 2nCn + 22n                                 

(Hence, proved)

(iii)    Let 0≤i<j≤n   Ci  Cj

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

                P = 0≤i<j≤n   (2n-i0j) Cn-i  Cn-j

                =  0≤i<j≤n   (2n-(i+j)) Ci Cj       [·.· nCn = nCn-r]

                = 2n 0≤i<j≤n   Ci Cj - P

        .·.     0≤i<j≤n   (i+j) Ci Cj = n[22n-1 - (2n)!/2(n!)2]                     

(Hence, proved)  

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