COT3100
Summer’2001
Recitation on induction (Solutions)
07/12/2001
1.
Use mathematical induction to prove that 3 divides
n
3
+2
n
whenever
n
is a
nonnegative integer. ,
We are going to prove by induction on
n
≥
0, that
n
3
+2
n
is divisible by 3.
Basis
.
n
=0. 0
3
+0=0 is divisible by 3, so the proposition holds for
n
=0.
IH
. Assume that the proposition is true for
n
=
k
, where
k
is some integer,
k
≥
0.
In other words we assume that
k
3
+2
k
is divisible by 3, i.e.
k
3
+2
k =
3
p
,
p
is an
integer.
IS
. We need to prove that (
k
+1)
3
+2(
k
+1) is divisible by 3.
(
k
+1)
3
+2(
k
+1) =
k
3
+3
k
2
+3
k
+1+2
k
+2=
= (
k
3
+2
k
) +3(
k
2
+
k
+1)
= 3
p
+3 (
k
2
+
k
+1)
by IH
Since both terms have factor 3, the sum is divisible by 3. So, induction step is proved.
By Induction Principle the proposition is true for any
n
.
2.
Use induction to prove that
3+3
⋅
5+3
⋅
5
2
+… +3
⋅
5
n
= 3(5
n
+1

1)/4
We are going to prove by induction on
n
≥
0 that the equality is true.
Basis
.
n
=0. LHS =3, RHS=3(5
0+1

1)/4=3. So, LHS=RHS.
IH
. Assume that the proposition is true for
n
=
k
, where
k
is some integer
k
≥
0. In other
words we assume that for some
k
≥
0 3+3
⋅
5+3
⋅
5
2
+… +3
⋅
5
k
= 3(5
k
+1

1)/4
IS
. We need to prove that 3+3
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 Spring '09
 Mathematical Induction, Recursion, Inductive Reasoning, Negative and nonnegative numbers, Natural number, Prime number

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