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Solved Subjective Question on Binomial Expression Set 1

Posted on - 04-01-2017

Math

IIT JEE

Example.1

Find the degree of the expression .

Solution

Since

=

=

\ Degree equal to 7.

Example.2

Show that there will be no term containing x2r in the expansion of (x + x-2)n-3 if n –2r is positive but not a multiple of 3.

Solution

If possible suppose that the kth term contains x2r in the expansion of
(x + x-2)n-3 , kth term is given by

=

since the kth term contains x2r \ 2r = n –3k ….(1).

or 3k = n –2r …..(2)

Now if n –2r is positive and not divisible by 3 then it is obvious from eqn. (2) that k will not be a positive integer.

Hence if (n –2r) is not divisible by 3 there will be no term containing x2r in the given expansion. .

Example.3

If x > 0 and the 4th term in the expansion of has maximum value then x > 64/21.

Solution

Since T4 is the greatest term in the expansion of ,

T3 < T4 and T5 < T4 < 1 and > 1

But

=

\ 1 > = .

Also 1 < = .

Thus x > 64/21.

Example.4

The coefficient of a10b7c3 in the expansion of ( bc + ca + ab)10 is

Solution

The general term in the expansion of ( bc + ca + ab)10 is

where r + s + t = 10

For the coefficient of a10b7c3 we set t + s = 10, r + t = 7, r + s = 3.

Since r + s + t = 10 we get r = 0, s = 3, t = 7. .

Thus, the coefficient of a10b7c3 in the expansion of ( bc + ca + ab)10 is

.

Example.5

Show that 32n+2 – 8n – 9 is divisible by 64, " n∈N.

Solution

32n+2 – 8n–9 = 32(32)n –8n–9 = 9(1+8)n–8n – 9

= 9

= 64 n +9 (nC282 + nC3.83 + -------- +8n).

= 64 [n+9(nC2+nC3.8------+8n-2)].

= 64K where K is an integer.

Hence 32n+2 – 8n–9 is divisible by 64.

 
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