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in the same file you will find all the problems
please solve the problem as the instructions need
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Explanation & Answer
The solutions are ready. Please read them and ask if something is unclear.I still have to polish some answers, those 3 hours you mentioned must be sufficient.docx and pdf files are identical
0. I hope we know that
𝑛
∑ 1 = 𝑛,
𝑖=1
𝑛
𝑛
∑𝑖 = ∑𝑖 =
𝑖=1
𝑛
𝑖=0
𝑛
(𝑛 + 1)𝑛
,
2
∑ 𝑖2 = ∑ 𝑖2 =
𝑖=1
𝑛
𝑖=0
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
,
6
2
(𝑛 + 1)𝑛
∑𝑖 = ∑𝑖 = (
) .
2
3
𝑖=1
3
𝑖=0
1.
𝑛
1) ⌊ 2⌋ (obvious).
𝑛
2) 𝑛 − ⌊ 2⌋ + 1 (obvious).
3) 𝑛 ∙ 2𝑛 = 𝟐𝒏𝟐 because the loops are independent.
4) (𝑛 − 2 + 1)(3𝑛) = 𝟑𝒏𝟐 − 𝟑𝒏 (the loops are independent).
5) Here the loops are dependent. We have
𝑛−1
𝑛−1
𝑛−1
∑(𝑛 − (𝑖 + 1) + 1) = ∑(𝑛 − 𝑖) = |𝑐ℎ𝑎𝑛𝑔𝑒 𝑖 → 𝑛 − 𝑖| = ∑ 𝑖 =
𝑖=1
𝑖=1
𝑖=1
(𝒏 − 𝟏)𝒏
.
𝟐
6) Similar to 5:
𝑛−1
𝑛−1
𝑛−1
∑(𝑛 − (𝑖) + 1) = ∑(𝑛 − 𝑖 + 1) = ∑(𝑖 + 1) =
𝑖=1
𝑖=1
𝑖=1
7) Similar to 5 and even simpler: ∑𝑛𝑖=1 𝑖 =
8) Similar to 7: ∑𝑛𝑖=1(𝑖 − 1) =
(𝑛+1)𝑛
2
−𝑛=
(𝒏+𝟏)𝒏
𝟐
.
(𝒏−𝟏)𝒏
𝟐
.
(𝑛 − 1)𝑛
(𝒏 − 𝟏)(𝒏 + 𝟐)
+𝑛−1=
.
2
𝟐
9) It is ∑𝑛𝑖=1 (⌊
and ⌊
𝑖+1
2
𝑖+1
2
⌋). Divide it into two sums, with odd and even 𝑖s. For an odd 𝑖, 𝑖 = 2𝑘 + 1
⌋ = 𝑘 + 1, for an even 𝑖, 𝑖 = 2𝑘 and ⌊
𝑛
∑ (⌊
𝑖=1
𝑖+1
2
𝑛−1
⌊
⌋
2
⌋ = 𝑘. This way the sum is
𝑛
⌊ ⌋
2
𝑖+1
⌋) = ∑ (𝑘 + 1) + ∑(𝑘).
2
𝑘=0
𝑘=1
We know what it is. We can consider odd and even 𝑛s separately and get nicer formulas.
10) almost the same as 9) (change of variable j).
𝑘
𝑚
𝑛
11) It is ∑𝑛𝑚=1(∑𝑚
𝑘=1(∑𝑗=1 𝑗)) = ∑𝑚=1 (∑𝑘=1 (
𝑛
𝑘(𝑘+1)
2
1
2
)) = 2 ∑𝑛𝑚=1(∑𝑚
𝑘=1(𝑘 + 𝑘)) =
𝑛
1
1
1
1
= ∑ ( (𝑛(𝑛 + 1)(2𝑛 + 1) + 𝑛(𝑛 + 1))) =
∑ (2𝑛3 + 3𝑛2 + 𝑛 + 3𝑛2 + 3𝑛) =
2
6
2
12
𝑚=1
𝑛
=
𝑚=1
1
1 1 2
∑ (2𝑛3 + 6𝑛2 + 4𝑛) =
( 𝑛 (𝑛 + 1)2 + 𝑛(𝑛 + 1)(2𝑛 + 1) + 2𝑛(𝑛 + 1)) =
12
12 2
𝑚=1
1
1
=
𝑛(𝑛 + 1)(𝑛(𝑛 + 1) + 2(2𝑛 + 1) + 4) =
𝑛(𝑛 + 1)(𝑛2 + 5𝑛 + 6) =
24
24
𝟏
=
𝒏(𝒏 + 𝟏)(𝒏 + 𝟐)(𝒏 + 𝟑).
𝟐𝟒
2.
1) 𝑖 may be any from 0 to 20, for any 𝑖 𝑗 may be any from 0 to 20 − 𝑖 and after 𝑖, 𝑗 are
chosen, 𝑘 is determined. So, the quantity is
20
20−𝑖
20
20
20
∑ ( ∑ (1)) = ∑(21 − 𝑖) = ∑ 21 − ∑ 𝑖 = 212 −
𝑖=0
𝑗=0
𝑖=0
𝑖=0
𝑖=0
20 ∙ 21
= 441 − 210 = 𝟐𝟑𝟏.
2
2) 𝑖 may be any from 1 to 20, for any 𝑖 𝑗 may be any from 1 to 20 − 𝑖 and after 𝑖, 𝑗 are
chosen, 𝑘 is determined. So, the quantity is
20
20−𝑖
20
20
20
∑ ( ∑ (1)) = ∑(20 − 𝑖) = ∑ 20 − ∑ 𝑖 = 202 −
𝑖=1
𝑗=1
𝑖=1
𝑖=1
𝑖=1
20 ∙ 21
= 400 − 210 = 𝟏𝟗𝟎.
2
3) Here we have the same way
30
30−𝑖
30−𝑖−𝑗
30
30−𝑖
30
∑ ( ∑ ( ∑ (1))) = ∑ ( ∑ (31 − 𝑖 − 𝑗)) = ∑ ((31 − 𝑖)2 −
𝑖=0
𝑗=0
𝑘=0
𝑖=0
𝑗=0
30
30
= |𝑖 → 30 − 𝑖| = ∑ ((𝑖 + 1)2 −
𝑖=0
30
𝑖=0
(30 − 𝑖)(31 − 𝑖)
)=
2
𝑖(𝑖 + 1)
1
) = ∑(2𝑖 2 + 4𝑖 + 2 − 𝑖 2 − 𝑖) =
2
2
𝑖=0
1
1 30 ∙ 31 ∙ 61 90 ∙ 31
1
∑(𝑖 2 + 3𝑖 + 2) = (
+
+ 62) = (350 ∙ 31 + 62) = 𝟓𝟒𝟓𝟔.
2
2
6
2
2
=
𝑖=0
4) It is similar to 3) but variables start from 2, not from 0:
30
30−𝑖
30−𝑖−𝑗
30
30−𝑖
∑ ( ∑ ( ∑ (1))) = ∑ ( ∑ (29 − 𝑖 − 𝑗)) =
𝑖=2
𝑗=2
𝑘=2
𝑖=2
30
= ∑ ((29 − 𝑖)2 − (
𝑖=2
𝑗=2
(30 − 𝑖)(31 − 𝑖)
− 1)) =
2
29
29
1
1
= |𝑖 → 31 − 𝑖| = ∑ ((𝑖 − 2) − (𝑖(𝑖 − 1) − 2)) = ∑(2𝑖 2 − 8𝑖 + 8 − 𝑖 2 + 𝑖 + 2) =
2
2
2
29
𝑖=1
𝑖=1
1
1 29 ∙ 30 ∙ 59 7 ∙ 29 ∙ 30
1
= ∑(𝑖 2 − 7𝑖 + 10) = (
−
+ 290) = (190 ∙ 29 + 290) = 29 ∙ 100 =
2
2
6
2
2
𝑖=1
= 𝟐𝟗𝟎𝟎.
500−𝑖−𝑗
500−𝑖
5) It is ∑500
𝑖=10(∑𝑗=10 (∑𝑘=10
490
480−𝑖
470−𝑖−𝑗
𝑗=0
490
480−𝑖
𝑖=0
𝑗=0
𝑘=0
(1)))) = |𝑖 → 𝑖 + 10, 𝑗 → 𝑗 + 10 𝑒𝑡𝑐| =
460−𝑖−𝑗−𝑘
= ∑( ∑ ( ∑ (
𝑖=0
500−𝑖−𝑗−𝑘
(∑𝑚=10
∑
490
480−𝑖
470−𝑖−𝑗
(1)))) = ∑ ( ∑ ( ∑ (461 − 𝑖 − 𝑗 − 𝑘))) =
𝑚=0
𝑖=0
𝑗=0
𝑘=0
1
= ∑ ( ∑ ((461 − 𝑖 − 𝑗)(471 − 𝑖 − 𝑗) − (470 − 𝑖 − 𝑗)(471 − 𝑖 − 𝑗))) =
2
490
480−𝑖
𝑖=0
𝑗=0
490
480−𝑖
𝑖=0
490
𝑗=0
1
= ∑ ( ∑ ((471 − 𝑖 − 𝑗)(452 − 𝑖 − 𝑗))) =
2
1
= ∑ ( ∑ ((471 − 𝑖)(452 − 𝑖) − 𝑗(923 − 2𝑖) + 𝑗 2 )) =
2
1
1
= ∑ ((481 − 𝑖)(471 − 𝑖)(452 − 𝑖) − (923 − 2𝑖) (480 − 𝑖)(481 − 𝑖)
2
2
𝑖=0
1
+ (480 − 𝑖)(481 − 𝑖)(961 − 2𝑖)) = ⋯
6
Here we need to change variables again and finish computations. Technology says the
answer is 1901790791.
It says the same about the initial expression and the last one, so we did not make any
errors😊
10−𝑖−𝑗
10−𝑖−𝑗−𝑘
10−𝑖
(1)))) = 𝟏𝟎𝟎𝟏
3. It is ∑10
(∑𝑚=0
𝑖=0(∑𝑗=0 (∑𝑘=0
(no doubt we can compute this the same way as above; I used technology here).
𝑖, 𝑗, 𝑘, 𝑚 correspond to the first 4 digits here.
𝑛+2
).
3
...