EECS 203: Homework 5 Solutions
Section 2.4
1. (E) 16bc
b)
∑
j
=
0
8
3
j
−
2
j
=
∑
j
=
0
8
3
j
−
∑
j
=
0
8
2
j
=
3
9
−
1
3
−
1
−
2
9
−
1
2
−
1
=
9841
−
511
=
9330
c)
∑
j
=
0
8
2
⋅
3
j
3
⋅
2
j
=
2
⋅
∑
j
=
0
8
3
j
3
⋅
∑
j
=
0
8
2
j
=
2
⋅
3
9
−
1
3
−
1
3
⋅
2
9
−
1
2
−
1
=
2
⋅
9841
3
⋅
511
=
21215
2. (E)
∑
i
=
1
n
∑
j
=
1
m
2
i
3
j
=
∑
i
=
1
n
∑
j
=
1
m
2
i
∑
i
=
1
n
∑
j
=
1
m
3
j
=
∑
i
=
1
n
2
mi
∑
i
=
1
n
3
⋅
m
⋅
m
1
2
=
2
m
⋅
n
⋅
n
1
2
3
n
⋅
m
⋅
m
1
2
=
n
⋅
m
⋅
2
n
3
m
5
2
3. (E)
∑
j
=
1
m
∑
i
=
1
n
2
i
3
j
=
∑
j
=
1
m
∑
i
=
1
n
2
i
∑
j
=
1
m
∑
i
=
1
n
3
j
=
∑
j
=
1
m
2
⋅
n
⋅
n
1
2
∑
j
=
1
m
3
n j
=
2
m
⋅
n
⋅
n
1
2
3
n
⋅
m
⋅
m
1
2
=
n
⋅
m
⋅
2
n
3
m
5
2
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View Full Document4. (M) 26
Solution 1:
The problem is to find a formula for
∑
k
=
0
m
⌊
3
k
⌋
.
The sum begins with (1+1+1+1+1+1+1)+(2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2)+(3+3+.
..+3)+.
...
The twos start with term 8. The threes start with term 27. In general, the number i starts with term i
3
.
Consider
subtracting one out of each element of the term. Then a total of m is subtracted, and the remaining terms are:
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 Spring '07
 YaoyunShi
 Addition, New interval

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