MAT 312/AMS 351, Summer 2011
Solutions to Homework Assignment 2
Maximal grade for HW2: 100 points
Section 1.2.
2. (15 points) Prove that for all positive integers
n
,
1 + 2
2
+ 3
2
+
...
+
n
2
=
n
(
n
+ 1)(2
n
+ 1)
6
.
(1)
Solution:
We prove this statement using induction by
n
.
Base.
We have to check the equation for
n
= 1:
1(1 + 1)(2
·
1 + 1)
6
=
2
·
3
6
= 1
.
Step.
Suppose that we proved the formula (1) for some number
n
, let us
prove it for
n
+ 1. We have:
1+2
2
+3
2
+
...
+
n
2
+(
n
+1)
2
= (1+2
2
+3
2
+
...
+
n
2
)+(
n
+1)
2
=
n
(
n
+ 1)(2
n
+ 1)
6
+(
n
+1)
2
=
n
+ 1
6
(
n
(2
n
+1)+6(
n
+1)) =
n
+ 1
6
(2
n
2
+7
n
+6) =
n
+ 1
6
(
n
+2)(2
n
+3) =
(
n
+ 1)((
n
+ 1) + 1)(2(
n
+ 1) + 1)
6
.
The last formula is precisely the right hand side of (1) with
n
replaced by
(
n
+ 1), so the equation (1) is true for (
n
+ 1) as well.
3. (15 points) The Fibonacci sequence is the sequence 1
,
1
,
2
,
3
,
5
,
8
,
13
...
where each term is the sum of the two preceding terms. Show that every two
successive terms of the Fibonacci sequence are relatively prime.
1
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View Full DocumentSolution:
We can deﬁne the Fibonacci sequence as follows:
F
1
=
F
2
= 1
,
F
n
=
F
n

1
+
F
n

2
for
n >
2
.
Let us prove that
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 Summer '08
 BESCHER
 Algebra, Remainder, Integers, Fibonacci number, summands, zp

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