Section 5.1
109
(h) For
n
=
1,
2
n
+
3
n

5
n
=
0 is divisible by 6. Now assume that
k
::::: 1 and that
2k
+
3
k

5
k
is
divisible by 6; that is, assume that
5
k
=
+
3
k
+
6N
for some integer
N.
We have
2k+l
+
3k+l
_
5
k
+
1
=
2(2k)
+
3(3
k
)

5(5
k
)
=
2(2k)
+
3(3
k
)

5(2k
+
3
k
+
6N)
using the induction hypothesis,
= 3(2k)

2(3
k
)

30N
=
_3(2)(2kl)

2(3)(3
k

1
)

30N,
which is divisible by 6 because each term is divisible by 6. By the Principle of Mathematical
Induction, we conclude that
2
n
+
3
n

5
n
is divisible by 6 for all
n
::::: 1.
(i) For
n
=
1,
16
n
+
IOn

1 = 25 is divisible by 25. Now assume that
k
::::: 1 and that
k
+
IOkl
is divisible by 25; that is, assume that
k
=
1  10k
+
25N
for some integer
We have
16k+l
+
lO(k
+
1)  1
=
16(16
k
)
+
10k
+
9
=
16(1 
10k
+
25N)
+
+
9
using the induction hypothesis,
=
25 
150k
+
16(25N),
which is divisible by 25 because each term is divisible by 25. By the Principle of Mathematical
Induction, we conclude that
n
+
IOn  1 is divisible by 25 for all
n
::::: 1.
(j)
[BB] For
n
=
1, 2!
=
2 is divisible by 2
1
,
so the result is true. Now assume that
k
::::: 1 and that
(2k)!
is divisible by
2k.
We have
[2(k
+
I)J!
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 Summer '10
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 Graph Theory, Natural number, 2 K, Mathematical logic, 3k, 3 k, 5 K

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