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Solved Subjective Question on Binomial Expression Set 2

Posted on - 04-01-2017

Math

IIT JEE

Example.1

If n is an in integer greater than 1, prove that

a–C1(a–1)+C2(a–2) – . . . + (–1)n Cn (a–n) = 0.

Solution

LHS = a –C1(a–1) + c2 (a–2) –. . .+ (–1)n Cn (a–n).

= a [1–C1+C2 – . . . (–1)n Cn] + [C1–2C2+3C3– . . .+(–1)n–1 nCn].

= aP + Q . . . .(1).

P = C0 –C1 + C2 – . . . + (–1)n Cn = 0 . . . (2).

Also,

C0 + C1x + C2x2 + . . . + Cnxn = (1+x)n.

Differentiating w.r.t. x,.

C1 + 2C2x + 3C3x2 + . . . . + nCnxn-1 = n(1+x)n–1.

Putting x = –1

C1 –2C2 + 3C3 –. . . + (–1)n–1 nCn= 0 . . . (3).

LHS = aP + Q = a´ 0 + 0 = 0. .

Example.2

If and ak = 1 for all k ≥n. Then show that
bn = 2n+1Cn + 1 .

Solution

Put x - 3 = y, we get . We have to find coefficient of yn in Left hand side as coefficient of yn in the right hand side is bn. And we will start getting coefficient of yn only when r ≥ n in the left hand side.

Coefficient of yn in (1 + y)n + (1 + y)n + 1 + ... + (1 + y)2n = bn.

Coefficient of yn in (1 + y)n


&⇒
Coefficient of yn + 1 in = bn
&⇒
2n+1Cn+1 = bn .

Example.3

If (1 + x)n = show that

Solution

rth factor of is given by

tr =

Now = t1 . t2 . t3 …… tn

=

=

=

Example.4

Prove that = (–1)n–1 nCn

Solution

(1+x)2n = 2nC0 + 2nC1x+2nC2x2 + . . . + 2nC2nx2n . . . (1).

Differentiating w.r.t. x, we get .

2n(1+x)2n–1 = C1 +2C2x + 3C3x2+ . . . +2nC2nx2n-1 . . . (2).

Also = C0 – C1 +C2 . . . + C2n . . . (3)

Where Cr = 2nCr

Multiplying (2) and (3), we get, 2n(1+x)2n–1

= (C1+2C2 x +3C3x2 + . . . +2nC2n x2n-1) (C0 –C1+C2. . . C2n )

The coefficient of in RHS is

=

Also, the coefficient of in 2n(1+x)2n-1

= coefficient of in 2n.

= coefficient of in (1–x2)2n–1(1–x)

= coefficient of x2n–2 in 2n (1–x2)2n–1(1–x)

= coefficient of x2n-1 in 2n (1–x2)2n–1 ´ coefficient of x in(1–x)

= coefficient of (x2)n–1 in 2n (1–x2)2n–1 ´ coefficient of x in (1–x)

= 2n (-1)n–1 2n–1Cn–1(-1) = (-1)n 2n

= (-1)n n = (-1)n n = - (- 1)n–1 nCn …. (6)

From (5) and (6), we get, = (-1)n–1 nCn

Example.5

Given sn = 1 + q +q2+........+qn.

Sn =

Prove that (n+1)C1 + (n+1)C2s1 + (n+1)C3s2+........+(n+1)Cn+1sn = 2n Sn.

Solution

sn is in geometric progression, hence

Consider (n+1)C1 + (n+1)C2s1 + (n+1)C3s3+........+(n+1)Cn+1sn.

=

Since

 
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