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Continue Shopping ```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 nCk (sin kx) cos (n-k)x = 2n-1 sin(nx)

Solution:

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

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

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

= 1/2 sin n x 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

=∫01 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 (-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 = 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 = (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 (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

=> n+1Cr+1 Sr=(1/(1-q)) n+1Cr+1- (1/(1-q)) n+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 Cr3  = 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 k3 [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

= (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-1 - (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) + 2∑0≤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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