a) (1+(−1)
j
j=0
8
∑
)=2 +0+2+0+2+0+2+0+2=10.
b) (3
j
j=0
8
∑
−2
j
) = 3
j
j=0
8
∑
− 2
j
j=0
8
∑
=3
9
−1
2−2
9
−1
1=9330.
c) (2⋅3
j
j=0
8
∑
+3⋅2
j
)=2⋅ 3
j
j=0
8
∑
+3⋅ 2
j
j=0
8
∑
=2⋅ 3
9
−1
2
⎡
⎣
⎢ ⎤
⎦
⎥ +3⋅ 2
9
−1
1
⎡
⎣
⎢ ⎤
⎦
⎥
=21215.
d) (2
j+1
j=0
8
∑
−2
j
) =2
9
−2
0
=511.
a) (i −j)
j+1
j=1
2
∑
= (i −1+i −2) = (2i −3) =(2−3)+(4−3) +(6−3) =3
i =1
3
∑
i =1
3
∑
i=1
3
∑
.
a
k
=−1
(k +1).
a
n
−a
0
=−1
(n +1) +1= n
n+1.
k
2
−(k −1)
2
[ ]
k=1
n
∑
= (2k −1)
k=1
n
∑
.
n2−0 = (2k −1)
k =1
n
∑,
CS1050 HW 6 Answer Key
There are 130 points total over the entire homework.
1) [5 points each]
Note: b) and c) below use the closed form for geometric series; d) below uses a
telescoping sum. (You could also use the geometric series form for d).)
2) [5 points each]
3) [20 points]
The rightmost side is now in the form of a telescoping sum with The result is
4) [20 points]
a) Find
Use the telescoping sum formula with ak = k2.
=(9⋅0+6) +(9⋅1+6) +(9⋅ 2+6) +(9⋅3+6) =78.
b) (3i +2j)
j=0
2
∑
= 3i +(3i +2)+(3i +4) = (9i +6)
i=0
3
∑
i=0
3
∑
i=0
3
∑
.
c) j
j=0
2
∑
= (0+1+2) = 3
i=1
3
∑
i =1
3
∑
i =1
3
∑
=9.
d) i
2
j
3
j=0
3
∑
= i
2
⋅0+i
2
⋅1+i
2
⋅8+i
2
⋅27
i=0
2
∑
i=0
2
∑
= 36i
2
i =0
2
∑
=(36⋅0 +36⋅1+36⋅4)=180.
1
k(k +1)
k=1
n
∑
=1
k−1
(k +1)
⎡
⎣
⎢ ⎤
⎦
⎥
k =1
n
∑
=−1
(k +1) −−1
k
⎡
⎣
⎢ ⎤
⎦
⎥
k=1
n
∑
.
