How did they get the term $(n+1)^3$ in the step of inductive proof which says $sum_k=1^n+1 k^3=sum_k=1^n k^3 + (n+1)^3$?
Clash Royale CLAN TAG#URR8PPP
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I'm struggling to understand on what was done in this inductive step. How did they get the $(n+1)^3$ term?
Proof Solution
summation induction
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
asked Jun 14 at 6:56
aqw143
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up vote
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I'm struggling to understand on what was done in this inductive step. How did they get the $(n+1)^3$ term?
Proof Solution
summation induction
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
asked Jun 14 at 6:56
aqw143
223
4
Do you see the indices of the summation? They change on the right hand side, because you are removing one term of the summation, namely $(n+1)^3$ and keeping it separately on the RHS. In other words, $sum_k=1^n+1 k^3$ is $1^3 + 2^3 + ... + (n+1)^3$, which is equal to $(1^3 + 2^3 + ... + n^3) + (n+1)^3$. Now the first part is the sum of cubes from $1$ to $n$, which is separately, and the $(n+1)^3$ is written separately.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jun 14 at 6:58
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm struggling to understand on what was done in this inductive step. How did they get the $(n+1)^3$ term?
Proof Solution
summation induction
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
asked Jun 14 at 6:56
aqw143
223
I'm struggling to understand on what was done in this inductive step. How did they get the $(n+1)^3$ term?
Proof Solution
summation induction
summation induction
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
asked Jun 14 at 6:56
aqw143
223
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
asked Jun 14 at 6:56
aqw143
223
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
edited Jun 14 at 10:04
Martin Sleziak
43.7k6113261
43.7k6113261
asked Jun 14 at 6:56
aqw143
223
asked Jun 14 at 6:56
aqw143
223
asked Jun 14 at 6:56
aqw143
223
223
4
Do you see the indices of the summation? They change on the right hand side, because you are removing one term of the summation, namely $(n+1)^3$ and keeping it separately on the RHS. In other words, $sum_k=1^n+1 k^3$ is $1^3 + 2^3 + ... + (n+1)^3$, which is equal to $(1^3 + 2^3 + ... + n^3) + (n+1)^3$. Now the first part is the sum of cubes from $1$ to $n$, which is separately, and the $(n+1)^3$ is written separately.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jun 14 at 6:58
add a comment |Â
4
Do you see the indices of the summation? They change on the right hand side, because you are removing one term of the summation, namely $(n+1)^3$ and keeping it separately on the RHS. In other words, $sum_k=1^n+1 k^3$ is $1^3 + 2^3 + ... + (n+1)^3$, which is equal to $(1^3 + 2^3 + ... + n^3) + (n+1)^3$. Now the first part is the sum of cubes from $1$ to $n$, which is separately, and the $(n+1)^3$ is written separately.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jun 14 at 6:58
4
4
Do you see the indices of the summation? They change on the right hand side, because you are removing one term of the summation, namely $(n+1)^3$ and keeping it separately on the RHS. In other words, $sum_k=1^n+1 k^3$ is $1^3 + 2^3 + ... + (n+1)^3$, which is equal to $(1^3 + 2^3 + ... + n^3) + (n+1)^3$. Now the first part is the sum of cubes from $1$ to $n$, which is separately, and the $(n+1)^3$ is written separately.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jun 14 at 6:58
Do you see the indices of the summation? They change on the right hand side, because you are removing one term of the summation, namely $(n+1)^3$ and keeping it separately on the RHS. In other words, $sum_k=1^n+1 k^3$ is $1^3 + 2^3 + ... + (n+1)^3$, which is equal to $(1^3 + 2^3 + ... + n^3) + (n+1)^3$. Now the first part is the sum of cubes from $1$ to $n$, which is separately, and the $(n+1)^3$ is written separately.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jun 14 at 6:58
add a comment |Â
5 Answers
5
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accepted
Underbrace to the rescue!
- $$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
- $$sum_k=1^n+1k^3=underbrace1^3+2^3+3^3+cdots+n^3_sum_k=1^nk^3+(n+1)^3$$
$$therefore, sum_k=1^n+1k^3=sum_k=1^nk^3+(n+1)^3$$
QED
answered Jun 14 at 7:21
BCLC
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$sumlimits_k=1^colorredn+1 k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + colorblue(n+1)^3=$
$[1^3+2^3 + 3^3 + ...... + n^3] + colorblue(n+1)^3 =$
$sumlimits_k=1^colorredn k^3 + colorblue(n+1)^3$
answered Jun 14 at 7:12
fleablood
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$$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
$$sum_k=1^n+1k^3=1^3+2^3+3^3+cdots+(n+1)^3=sum_k=1^nk^3+(n+1)^3$$
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
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up vote
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In the step first step, the sum is from $k=1$ to $n+1$, so the last term is $(n+1)^3.$
in the next step, he added $(n+1)^3$ outside of the sum, and removed the last term in the sum by changing $n+1$ to $n.$
answered Jun 14 at 7:02
Prog R. Hammer
338312
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here k=n+1
$$sum_i=0^n i^3 = 1^3+2^3+3^3+.....+n^3$$
$$sum_i=0^k i^3 = 1^3+2^3+3^3+.....+n^3+(n+1)^3$$
thus $$sum_i=0^k i^3$$=$$sum_i=0^n i^3$$+$$(n+1)^3$$
answered Jun 14 at 7:20
user180165
254
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Underbrace to the rescue!
- $$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
- $$sum_k=1^n+1k^3=underbrace1^3+2^3+3^3+cdots+n^3_sum_k=1^nk^3+(n+1)^3$$
$$therefore, sum_k=1^n+1k^3=sum_k=1^nk^3+(n+1)^3$$
QED
answered Jun 14 at 7:21
BCLC
1
add a comment |Â
up vote
8
down vote
accepted
Underbrace to the rescue!
- $$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
- $$sum_k=1^n+1k^3=underbrace1^3+2^3+3^3+cdots+n^3_sum_k=1^nk^3+(n+1)^3$$
$$therefore, sum_k=1^n+1k^3=sum_k=1^nk^3+(n+1)^3$$
QED
answered Jun 14 at 7:21
BCLC
1
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Underbrace to the rescue!
- $$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
- $$sum_k=1^n+1k^3=underbrace1^3+2^3+3^3+cdots+n^3_sum_k=1^nk^3+(n+1)^3$$
$$therefore, sum_k=1^n+1k^3=sum_k=1^nk^3+(n+1)^3$$
QED
answered Jun 14 at 7:21
BCLC
1
Underbrace to the rescue!
- $$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
- $$sum_k=1^n+1k^3=underbrace1^3+2^3+3^3+cdots+n^3_sum_k=1^nk^3+(n+1)^3$$
$$therefore, sum_k=1^n+1k^3=sum_k=1^nk^3+(n+1)^3$$
QED
answered Jun 14 at 7:21
BCLC
1
edited Sep 8 at 10:41
answered Jun 14 at 7:21
BCLC
1
answered Jun 14 at 7:21
BCLC
1
answered Jun 14 at 7:21
BCLC
1
1
add a comment |Â
add a comment |Â
up vote
2
down vote
$sumlimits_k=1^colorredn+1 k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + colorblue(n+1)^3=$
$[1^3+2^3 + 3^3 + ...... + n^3] + colorblue(n+1)^3 =$
$sumlimits_k=1^colorredn k^3 + colorblue(n+1)^3$
answered Jun 14 at 7:12
fleablood
61.7k22678
add a comment |Â
up vote
2
down vote
$sumlimits_k=1^colorredn+1 k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + colorblue(n+1)^3=$
$[1^3+2^3 + 3^3 + ...... + n^3] + colorblue(n+1)^3 =$
$sumlimits_k=1^colorredn k^3 + colorblue(n+1)^3$
answered Jun 14 at 7:12
fleablood
61.7k22678
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$sumlimits_k=1^colorredn+1 k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + colorblue(n+1)^3=$
$[1^3+2^3 + 3^3 + ...... + n^3] + colorblue(n+1)^3 =$
$sumlimits_k=1^colorredn k^3 + colorblue(n+1)^3$
answered Jun 14 at 7:12
fleablood
61.7k22678
$sumlimits_k=1^colorredn+1 k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + colorblue(n+1)^3=$
$[1^3+2^3 + 3^3 + ...... + n^3] + colorblue(n+1)^3 =$
$sumlimits_k=1^colorredn k^3 + colorblue(n+1)^3$
answered Jun 14 at 7:12
fleablood
61.7k22678
answered Jun 14 at 7:12
fleablood
61.7k22678
answered Jun 14 at 7:12
fleablood
61.7k22678
answered Jun 14 at 7:12
fleablood
61.7k22678
61.7k22678
add a comment |Â
add a comment |Â
up vote
1
down vote
$$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
$$sum_k=1^n+1k^3=1^3+2^3+3^3+cdots+(n+1)^3=sum_k=1^nk^3+(n+1)^3$$
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
add a comment |Â
up vote
1
down vote
$$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
$$sum_k=1^n+1k^3=1^3+2^3+3^3+cdots+(n+1)^3=sum_k=1^nk^3+(n+1)^3$$
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
$$sum_k=1^n+1k^3=1^3+2^3+3^3+cdots+(n+1)^3=sum_k=1^nk^3+(n+1)^3$$
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
$$sum_k=1^nk^3=1^3+2^3+3^3+cdots+n^3$$
$$sum_k=1^n+1k^3=1^3+2^3+3^3+cdots+(n+1)^3=sum_k=1^nk^3+(n+1)^3$$
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
answered Jun 14 at 7:03
Rhys Hughes
4,1081227
4,1081227
add a comment |Â
add a comment |Â
up vote
0
down vote
In the step first step, the sum is from $k=1$ to $n+1$, so the last term is $(n+1)^3.$
in the next step, he added $(n+1)^3$ outside of the sum, and removed the last term in the sum by changing $n+1$ to $n.$
answered Jun 14 at 7:02
Prog R. Hammer
338312
add a comment |Â
up vote
0
down vote
In the step first step, the sum is from $k=1$ to $n+1$, so the last term is $(n+1)^3.$
in the next step, he added $(n+1)^3$ outside of the sum, and removed the last term in the sum by changing $n+1$ to $n.$
answered Jun 14 at 7:02
Prog R. Hammer
338312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In the step first step, the sum is from $k=1$ to $n+1$, so the last term is $(n+1)^3.$
in the next step, he added $(n+1)^3$ outside of the sum, and removed the last term in the sum by changing $n+1$ to $n.$
answered Jun 14 at 7:02
Prog R. Hammer
338312
In the step first step, the sum is from $k=1$ to $n+1$, so the last term is $(n+1)^3.$
in the next step, he added $(n+1)^3$ outside of the sum, and removed the last term in the sum by changing $n+1$ to $n.$
answered Jun 14 at 7:02
Prog R. Hammer
338312
answered Jun 14 at 7:02
Prog R. Hammer
338312
answered Jun 14 at 7:02
Prog R. Hammer
338312
answered Jun 14 at 7:02
Prog R. Hammer
338312
338312
add a comment |Â
add a comment |Â
up vote
0
down vote
here k=n+1
$$sum_i=0^n i^3 = 1^3+2^3+3^3+.....+n^3$$
$$sum_i=0^k i^3 = 1^3+2^3+3^3+.....+n^3+(n+1)^3$$
thus $$sum_i=0^k i^3$$=$$sum_i=0^n i^3$$+$$(n+1)^3$$
answered Jun 14 at 7:20
user180165
254
add a comment |Â
up vote
0
down vote
here k=n+1
$$sum_i=0^n i^3 = 1^3+2^3+3^3+.....+n^3$$
$$sum_i=0^k i^3 = 1^3+2^3+3^3+.....+n^3+(n+1)^3$$
thus $$sum_i=0^k i^3$$=$$sum_i=0^n i^3$$+$$(n+1)^3$$
answered Jun 14 at 7:20
user180165
254
add a comment |Â
up vote
0
down vote
up vote
0
down vote
here k=n+1
$$sum_i=0^n i^3 = 1^3+2^3+3^3+.....+n^3$$
$$sum_i=0^k i^3 = 1^3+2^3+3^3+.....+n^3+(n+1)^3$$
thus $$sum_i=0^k i^3$$=$$sum_i=0^n i^3$$+$$(n+1)^3$$
answered Jun 14 at 7:20
user180165
254
here k=n+1
$$sum_i=0^n i^3 = 1^3+2^3+3^3+.....+n^3$$
$$sum_i=0^k i^3 = 1^3+2^3+3^3+.....+n^3+(n+1)^3$$
thus $$sum_i=0^k i^3$$=$$sum_i=0^n i^3$$+$$(n+1)^3$$
answered Jun 14 at 7:20
user180165
254
answered Jun 14 at 7:20
user180165
254
answered Jun 14 at 7:20
user180165
254
answered Jun 14 at 7:20
user180165
254
254
add a comment |Â
add a comment |Â
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4
Do you see the indices of the summation? They change on the right hand side, because you are removing one term of the summation, namely $(n+1)^3$ and keeping it separately on the RHS. In other words, $sum_k=1^n+1 k^3$ is $1^3 + 2^3 + ... + (n+1)^3$, which is equal to $(1^3 + 2^3 + ... + n^3) + (n+1)^3$. Now the first part is the sum of cubes from $1$ to $n$, which is separately, and the $(n+1)^3$ is written separately.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jun 14 at 6:58