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.
