1. The document proves through mathematical induction that the formula nj=1 j^2 = n(n+1)(2n+1)/6 for the sum of squares from 1 to n is true for all positive integers n. 2. It also proves that a 2n x 2n chessboard with one square missing can be covered with L-shaped pieces that each cover three squares, again using mathematical induction. 3. Mathematical induction is used to prove both formulas by showing the base case holds and assuming the formula is true for an integer n to prove it is true for n+1.

1. Student ID: U10011024 Name: Kuan-Lun Wang
Exercises 1.3-7
Use the identity (j +1)3 −j 3 = 3j 2 +3j +1 and mathematical
induction to prove that n j 2 = 12 + 22 + 32 + · · · + n2 =
j=1
n(n + 1)(2n + 1)/6 for every positive integer n.
(1)By the identity (j + 1)3 − j 3 = 3j 2 + 3j + 1.
When we isolate the factor j 2, we find that j 2 = ((j +
1)3 − j 3)/3 − j − 1/3). When we sum this expression for j
over the values j = 1, 2, . . . n, we obtain
n
tn = j2
j=1
 
n n n
3 3
=  ((j + 1) − j )/3 − j− 1/3
j=1 j=1 j=1
(replacing j 2 with ((j + 1) − j 3)/3 − j − 1/3)
3
= ((n + 1)3/3 − 1/3) − n(n + 1)/2 − n/3
(simplif ying a telescoping sum)
= n(n + 1)(2n + 1)/6.
This completes the proof by the identity (j + 1)3 − j 3 =
3j 2 + 3j + 1.
(2)By mathematical induction.
We can show by mathematical induction that n j 2 =
j=1
n(n + 1)(2n + 1)/6 for every positive integer n. The basis
step, namely, the case where n = 1, hold because 1 j 2 =
j=1
n
2
1 = 1(1 + 1)(2 × 1 + 1)/6. Now, assume that j=1 j 2 =
n(n + 1)(2n + 1)/6; this is the inductive hypothesis. To
complete the proof, we must show, under the assumption
that the inductive hypothesis is true, that n+1 j 2 = (n +
j=1
1)((n+1)+1)(2(n+1)+1)/6. Using the inductive hypothesis,
Home Work 02 1

2. Student ID: U10011024 Name: Kuan-Lun Wang
we have
n+1 n
j2 = j 2 + (n + 1)2
j=1 j=1
= n(n + 1)(2n + 1)/6 + (k 2 + 2k + 1)
= (n + 1)((n + 1) + 1)(2(n + 1) + 1)/6
This completes both the inductive step and the proof.
Exercises 1.3-34
Use mathematical induction to show that a 2n ×2n chessboard
with one square missing can be covered with L-shaped pieces,
where each L-shaped piece covers three square.
We can show by mathematical induction that a 2n × 2n
chessboard with one square missing can be covered with L-
shaped pieces, where each L-shaped piece covers three square.
The basis step, namely, the case where n = 1, hold because
a 21 × 21 chessboard with one square missing can be covered
with L-shaped pieces (since there are three squares remain-
ing, in each each case the chessboard can be tiled with a
L-shaped).
Now, assume that a 2n × 2n chessboard with one square
missing can be covered with L-shaped pieces; this is the in-
ductive hypothesis. To complete the proof, we must show,
under the assumption that the inductive hypothesis is true,
that a 2n+1 × 2n+1 chessboard with one square missing can
be covered with L-shaped pieces. We divide the 2n+1 × 2n+1
chessboard into four 2n ×2n chessboards. Using the inductive
hypothesis, we can tile the remaining three 2n × 2n chess-
boards so that each tiling omits the square closest to the cen-
ter of the 2n+1 × 2n+1 chessboard. Lest, place an L-shaped
Home Work 02 2

3. Student ID: U10011024 Name: Kuan-Lun Wang
piece in the center of the board such that it coves the corner
square of each remaining quadrant.
Now, we completes both the inductive step and the proof.
Home Work 02 3