5. Find the ratio of the coefficient of x*4 to the coefficient of x*5 in the expansion of (2-3x)*8
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The general term in the expansion of (2−x+3x
2
)
6
is
r!s!t!
6!
2
r
(−x)
s
(3x
2
)
t
, where r+s+t=6
=
r!s!t!
6
2
r
×(−1)
s
×3
t
×x
s+2t
,
where r+s+t=6
For the coefficient of x
5
, we must have
s+2t=5
But r+s+t=6
∴s=5−2t and r=1+t
where, 0≤r,s,t≤6
Now, t=0⇒r=1,s=5
t=1⇒r=2,s=3
t=2⇒r=3,s=1
Thus, there are three terms containing x
5
and coefficient of x
5
=
1!5!0!
6!
×2
1
×(−1)
5
×3
0
+
2!3!1!
6!
×2
2
×(−1)
3
×3
1
+
3!1!2!
6!
×2
3
×(−1)
1
×3
2
=−12−720−4320
=−5052.
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