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Solved Subjective Question on Binomial Expression Set 5

Posted on - 04-01-2017

Math

IIT JEE

Example.1

Prove that + + + -------- + = .

Solution

LHS = + + + -------- +

++ 2 + -------- + n

= n + (n–1) + (n–2) + …. …..(1).

= = RHS

Example.2

Find the sum of the series

Solution

The given series

Now, =

Similarly, etc.

Hence, the gives series =

= =

Example.3

If , prove that for n ≥ 2, .

Solution

For x ∈ R, we have

…. (1)

We know that ax2 + 2bx + c ≥ 0 for each x ∈ R if b2 –ac £ 0 and a > 0.

Since 2n –1 > 0, The inequality in (1) will hold if and only if

Example.4

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

(a)

(b)

Solution

(a) We have


&⇒

Differentiating both the sides we get

Putting x = 1 in the above expression, we get

…(1)

But

= …(2)

Let us now evaluate

We have (1+x)n = C0+C1x+C2x2 + …+Cnxn …(3)

Differentiating both sides w.r.t.x we get.


&⇒
…(4)

Replacing x by 1/x in (3) we get

…(5)

Note that = coefficient of constant term in

[using (4) and (5)]

= coefficient of constant term in

= coefficient of xn-1 in (1+x)n [n(1+x)n-1]

= coefficient of xn-1 in n (1+x)2n-1 = n. (2n-1Cn-1).

Thus, from (2) we get

=


&⇒


&⇒

(b) We have

=

Since

=


&⇒
2

= n

= = 2


&⇒
=

\

= + 2

Example.5

Show that nC0 –nC1+nC2+ . . . .

Solution

We have

=

+ . . . (1)

We know that

Differentiating both sides w.r.t y, we get.

-

Thus

= . . . .(2)

and -

= - . . . . (3)

Putting these values in (1), we obtain

= =

 
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