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154
Solutions to Exercises
Therefore,
f(x)
+
10xf(x)
+
25x
2
=
ao
+
(al
+
lOao)x
+
(a2
+
lOal
+
25ao)x2
+ .
.. +
(an
+
lOanl
+
25an_2)X
n
+ .
...
So
f(x)(l
+
lOx
+
2
)
=
1
+
35x.
Therefore,
1
= (1
+
5X)2
(1
+
35x)
=
+
2(
5x)
+ .
.. +
(n
+
1)(
5xt
+ .
..
)(1
+
= 1
+
25x

275x
2
+ .
.. +
[(n
+
1)(
5t
+
n(
5t
1
(35)]xn
+ .
= 1
+

2
+ .
.. +
[(1 
6n)( _5)n]xn
+ .
..
and so
an
=
(1 
6n)( _5)n.
11. [BB]
+
alX
+
a2x2
+
a3x3
+
+
anx
n
+
+
2a
n
_lX
n
+
+
an_2
Xn
+
+
2an_3
+
2xf(x)
2aox
+
2alX2
+
2a2x3
+
x
2
aox
2
+
alx
3
+
2x
3
f(x 2aox3
+
Therefore,


x
2
+
3
=
+
(al 
2ao)x
+
 2al 
ao)x
2
+
(a3

2a2
 al
+
2ao)x3
+ .
.. +
(an

2an
l

an2
+
2an_3)X
n
+ .
..
= 1
+
x

x
2
since
=
1, al 
2ao
=
3 
2(1)
=
1,
a2
=
6 
2(3)  1
=
1 and
an

2an
l 
+
2an3
=
0 for
n
~
3. So
and
f
1
+
x

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 Summer '10
 any
 Graph Theory

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