1. Write the following using summation symbol(s). (a) a3 + a4 + a5 + . . . + an−1 (b) a2 + a4 +?

... a6 + a8 + a10 + . . . + a2n(c) a3 + a5 + a7 + . . .(d) c9 + c8 + c7 + · · · + c2 + c1(e) 1 + 12 +14 +18 + . . .(f) n + (n &m

... a6 + a8 + a10 + . . . + a2n
(c) a3 + a5 + a7 + . . .
(d) c9 + c8 + c7 + · · · + c2 + c1
(e) 1 + 1
2 +
1
4 +
1
8 + . . .
(f) n + (n − 1) + (n − 2) + · · · + 2 + 1
(g) −1 + 2 − 3 + 4 − 5 + . . .
(h) ai + ai+2 + ai+4 + ai+6 + ai+8 + . . . + ai+2n
(i) 1
n +
2
n+1 +
3
n+2 + . . . +
n+1
2n
2. Given that Pn
i=1 i =
n(n+1)
2
, how can you evaluate Pn
i=34 i?
3. Evaluate P100
j=40 j
4. Evaluate P1000
10 3
5. Expand Pn
i=0
(−1)i
i+1
6. Expand and simplify P5
i=1(ai − ai−1)
7. Rewrite Pn+1
i=1
1
i
2 by separating off the last term.
8. Rewrite Pn
k=0 2
k + 2n+1 as a single summation.
9. We sometimes can simplify summation by changing the variable of summation. For
example, suppose P6
k=0
1
k+1 . Let us make the variable change, j = k + 1. Basically
we need to replace k + 1 or k whenever they occur. Systematically, we compute first
the lower bound of the summation, followed by the upper bound of the summation,
and lastly the term of the summation. In our case the lower bound is k = 0, hence
in that case j = k + 1 = 0 + 1 = 1; the upper bound is k = 6, and therefor the new
upper bound is j = k + 1 = 6 + 1 = 7. The term 1
k+1 becomes 1
j
. Hence we have
P6
k=0
1
k+1 =
P7
j=1
1
j
. The new summation looks a lot better.
1
(a) Express Pn−1
i=1
i
(n−i)
2 as a new summation by using j = i − 1 to change the summation
variable.
(b) Express Pn
i=3
i
i+n−1
as a new summation by using j = i − 1 to change the summation
variable.
10. Let a1 = −2, a2 = −1, a3 = 0, a4 = 1, and a5 = 2. Compute the following:
(a) P5
i=1 ai
(b) P2
i=0 a2i+1
11. Rewrite Pn+1
i=1
1
i
2 by separating off the last three terms.
12. Rewrite Pn
k=0 2
k by separating off the first and the last terms.
13. For each integer n with n ≥ 1, let P(n) be the formula:
1 + 3 + 5 + . . . + (2n − 1) = n
2
(a) Write 1 + 3 + 5 + . . . + (2n − 1) using a summation symbol.
(b) Using mathematical induction, prove that: summation you derived above = n
2
,
∀n ≥ 1. Use the template we used in CSI131.
14. Using mathematical induction, prove that: Pn
i=1(3i − 2) = n(3n−1)
2
, ∀n ≥ 1. Use the
template we used in CSI131.
15. Using mathematical induction, prove that: Pn
i=1(
1
i(i+1) ) = n
n+1 , ∀n ≥ 1. Use the
template we used in CSI131

or:... a6 + a8 + a10 + . . . + a2n(c) a3 + a5 + a7 + . . .(d) c9 + c8 + c7 + \u00b7 \u00b7 \u00b7 + c2 + c1(e) 1 + 12 +14 +18 + . . .(f) n + (n \u2212 1) + (n \u2212 2) + \u00b7 \u00b7 \u00b7 + 2 + 1(g) \u22121 + 2 \u2212 3 + 4 \u2212 5 + . . .(h) ai + ai+2 + ai+4 + ai+6 + ai+8 + . . . + ai+2n(i) 1n +2n+1 +3n+2 + . . . +n+12n2. Given that Pni=1 i =n(n+1)2, how can you evaluate Pni=34 i?3. Evaluate P100j=40 j4. Evaluate P100010 35. Expand Pni=0(\u22121)ii+16. Expand and simplify P5i=1(ai \u2212 ai\u22121)7. Rewrite Pn+1i=11i2 by separating off the last term.8. Rewrite Pnk=0 2k + 2n+1 as a single summation.9. We sometimes can simplify summation by changing the variable of summation. Forexample, suppose P6k=01k+1 . Let us make the variable change, j = k + 1. Basicallywe need to replace k + 1 or k whenever they occur. Systematically, we compute firstthe lower bound of the summation, followed by the upper bound of the summation,and lastly the term of the summation. In our case the lower bound is k = 0, hencein that case j = k + 1 = 0 + 1 = 1; the upper bound is k = 6, and therefor the newupper bound is j = k + 1 = 6 + 1 = 7. The term 1k+1 becomes 1j. Hence we haveP6k=01k+1 =P7j=11j. The new summation looks a lot better.1(a) Express Pn\u22121i=1i(n\u2212i)2 as a new summation by using j = i \u2212 1 to change the summationvariable.(b) Express Pni=3ii+n\u22121as a new summation by using j = i \u2212 1 to change the summationvariable.10. Let a1 = \u22122, a2 = \u22121, a3 = 0, a4 = 1, and a5 = 2. Compute the following:(a) P5i=1 ai(b) P2i=0 a2i+111. Rewrite Pn+1i=11i2 by separating off the last three terms.12. Rewrite Pnk=0 2k by separating off the first and the last terms.13. For each integer n with n \u2265 1, let P(n) be the formula:1 + 3 + 5 + . . . + (2n \u2212 1) = n2(a) Write 1 + 3 + 5 + . . . + (2n \u2212 1) using a summation symbol.(b) Using mathematical induction, prove that: summation you derived above = n2,\u2200n \u2265 1. Use the template we used in CSI131.14. Using mathematical induction, prove that: Pni=1(3i \u2212 2) = n(3n\u22121)2, \u2200n \u2265 1. Use thetemplate we used in CSI131.15. Using mathematical induction, prove that: Pni=1(1i(i+1) ) = nn+1 , \u2200n \u2265 1. Use thetemplate we used in CSI131

Tags:write,