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How to find the sum of the the first n elements in the series 1,2,4,5,7,8… as a function of n?
Clash Royale CLAN TAG#URR8PPP
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5
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I see I can split it up into two arithmetic series (1,4,7... and 2,5,8...) and add the two sums, but what is the compact notation for the sum of these two series as a function of n?
arithmetic-progressions
asked Sep 2 at 3:44
Ido Sarig
1313
add a comment |Â
up vote
5
down vote
favorite
I see I can split it up into two arithmetic series (1,4,7... and 2,5,8...) and add the two sums, but what is the compact notation for the sum of these two series as a function of n?
arithmetic-progressions
asked Sep 2 at 3:44
Ido Sarig
1313
9
You should define your sequence clearly. I assume it's just the naturals less the multiples of $3$ but why make your readers guess? If I am correct, the answer is then the difference between a triangular number and three times another triangular number.
– lulu
Sep 2 at 3:50
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I see I can split it up into two arithmetic series (1,4,7... and 2,5,8...) and add the two sums, but what is the compact notation for the sum of these two series as a function of n?
arithmetic-progressions
asked Sep 2 at 3:44
Ido Sarig
1313
I see I can split it up into two arithmetic series (1,4,7... and 2,5,8...) and add the two sums, but what is the compact notation for the sum of these two series as a function of n?
arithmetic-progressions
asked Sep 2 at 3:44
Ido Sarig
1313
asked Sep 2 at 3:44
Ido Sarig
1313
asked Sep 2 at 3:44
Ido Sarig
1313
asked Sep 2 at 3:44
Ido Sarig
1313
1313
9
You should define your sequence clearly. I assume it's just the naturals less the multiples of $3$ but why make your readers guess? If I am correct, the answer is then the difference between a triangular number and three times another triangular number.
– lulu
Sep 2 at 3:50
add a comment |Â
9
You should define your sequence clearly. I assume it's just the naturals less the multiples of $3$ but why make your readers guess? If I am correct, the answer is then the difference between a triangular number and three times another triangular number.
– lulu
Sep 2 at 3:50
9
9
You should define your sequence clearly. I assume it's just the naturals less the multiples of $3$ but why make your readers guess? If I am correct, the answer is then the difference between a triangular number and three times another triangular number.
– lulu
Sep 2 at 3:50
You should define your sequence clearly. I assume it's just the naturals less the multiples of $3$ but why make your readers guess? If I am correct, the answer is then the difference between a triangular number and three times another triangular number.
– lulu
Sep 2 at 3:50
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3fracm(m-1)2 + 2m + 3fracm(m-1)2 = 3 (m + m(m-1)) = frac3n^24.
$$
If $n$ is odd, then $n-1$ is even and we know that the sum of the first $n-1$ entries is $frac3(n-1)^24$. Moreover, the last entry in the sequence was the $((n-1)/2 + 1)$-st entry of the first arithmetic progression, that is $1,4,7,10, ldots$. But the value of that entry is $3 ((n-1)/2 + 1) - 2 = 3(n-1)/2 + 1$, so using the formula for the even case, the sum of the first $n$ entries equals
$$
frac3(n-1)^24 + 3(n-1)/2 + 1 = frac34 (n^2 - 2n + 2n + 1 - 2 + 4/3) = frac3n^24 + frac14.
$$
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
1
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
1
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
add a comment |Â
up vote
9
down vote
Your sequence $1,2,4,5,7,8,dots$ satisfies the second order nonhomogeneous linear recurrence
$$a_n=a_n-2+3$$
with initial values
$$a_1=1, a_2=2.$$
The solution is
$$a_n=frac6n-3-(-1)^n4.$$
The sum of the first $n$ terms is
$$sum_k=1^n a_k=frac34n^2+frac1-(-1)^n8=boxedfrac34n^2+frac1-cos npi8.$$
answered Sep 2 at 4:08
bof
46.7k349113
1
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
4
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
1
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
2
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
 |Â
show 3 more comments
up vote
4
down vote
If you consider
$$a_k=1+3kqquad b_k=2+3k$$ then
$$sum_k=0^n a_k=sum_k=0^n (1+3k)=sum_k=0^n 1+3sum_k=0^n k=n+1+3 frac12 n (n+1)=frac12 (n+1) (3 n+2)$$
$$sum_k=0^n b_k=sum_k=0^n (2+3k)=2sum_k=0^n 1+3sum_k=0^n k=2(n+1)+3 frac12 n (n+1)=frac12 (n+1) (3 n+4)$$
Just continue.
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3fracm(m-1)2 + 2m + 3fracm(m-1)2 = 3 (m + m(m-1)) = frac3n^24.
$$
If $n$ is odd, then $n-1$ is even and we know that the sum of the first $n-1$ entries is $frac3(n-1)^24$. Moreover, the last entry in the sequence was the $((n-1)/2 + 1)$-st entry of the first arithmetic progression, that is $1,4,7,10, ldots$. But the value of that entry is $3 ((n-1)/2 + 1) - 2 = 3(n-1)/2 + 1$, so using the formula for the even case, the sum of the first $n$ entries equals
$$
frac3(n-1)^24 + 3(n-1)/2 + 1 = frac34 (n^2 - 2n + 2n + 1 - 2 + 4/3) = frac3n^24 + frac14.
$$
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
1
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
1
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
add a comment |Â
up vote
2
down vote
accepted
Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3fracm(m-1)2 + 2m + 3fracm(m-1)2 = 3 (m + m(m-1)) = frac3n^24.
$$
If $n$ is odd, then $n-1$ is even and we know that the sum of the first $n-1$ entries is $frac3(n-1)^24$. Moreover, the last entry in the sequence was the $((n-1)/2 + 1)$-st entry of the first arithmetic progression, that is $1,4,7,10, ldots$. But the value of that entry is $3 ((n-1)/2 + 1) - 2 = 3(n-1)/2 + 1$, so using the formula for the even case, the sum of the first $n$ entries equals
$$
frac3(n-1)^24 + 3(n-1)/2 + 1 = frac34 (n^2 - 2n + 2n + 1 - 2 + 4/3) = frac3n^24 + frac14.
$$
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
1
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
1
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3fracm(m-1)2 + 2m + 3fracm(m-1)2 = 3 (m + m(m-1)) = frac3n^24.
$$
If $n$ is odd, then $n-1$ is even and we know that the sum of the first $n-1$ entries is $frac3(n-1)^24$. Moreover, the last entry in the sequence was the $((n-1)/2 + 1)$-st entry of the first arithmetic progression, that is $1,4,7,10, ldots$. But the value of that entry is $3 ((n-1)/2 + 1) - 2 = 3(n-1)/2 + 1$, so using the formula for the even case, the sum of the first $n$ entries equals
$$
frac3(n-1)^24 + 3(n-1)/2 + 1 = frac34 (n^2 - 2n + 2n + 1 - 2 + 4/3) = frac3n^24 + frac14.
$$
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
Let us first find the formula for even $n$. In this case, $m = n/2$ is a whole number, and moreover the sum of the first $n$ elements of your sequence is the sum of the first $m$ elements of the two arithmetic progressions, that is,
$$
m + 3fracm(m-1)2 + 2m + 3fracm(m-1)2 = 3 (m + m(m-1)) = frac3n^24.
$$
If $n$ is odd, then $n-1$ is even and we know that the sum of the first $n-1$ entries is $frac3(n-1)^24$. Moreover, the last entry in the sequence was the $((n-1)/2 + 1)$-st entry of the first arithmetic progression, that is $1,4,7,10, ldots$. But the value of that entry is $3 ((n-1)/2 + 1) - 2 = 3(n-1)/2 + 1$, so using the formula for the even case, the sum of the first $n$ entries equals
$$
frac3(n-1)^24 + 3(n-1)/2 + 1 = frac34 (n^2 - 2n + 2n + 1 - 2 + 4/3) = frac3n^24 + frac14.
$$
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
edited Sep 2 at 18:36
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
answered Sep 2 at 5:28
AlgebraicsAnonymous
1,04512
1,04512
1
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
1
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
add a comment |Â
1
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
1
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
1
1
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
Thanks, I get it, But shouldn't that sentence read "If n is odd, then n−1 is even and we know that the sum of the first n−1 entries is [3⋅(n−1)]^2/4?.
– Ido Sarig
Sep 2 at 18:33
1
1
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
You're completely right. I don't know why I wrote that. Reloading the page should yield the corrected version.
– AlgebraicsAnonymous
Sep 2 at 18:36
add a comment |Â
up vote
9
down vote
Your sequence $1,2,4,5,7,8,dots$ satisfies the second order nonhomogeneous linear recurrence
$$a_n=a_n-2+3$$
with initial values
$$a_1=1, a_2=2.$$
The solution is
$$a_n=frac6n-3-(-1)^n4.$$
The sum of the first $n$ terms is
$$sum_k=1^n a_k=frac34n^2+frac1-(-1)^n8=boxedfrac34n^2+frac1-cos npi8.$$
answered Sep 2 at 4:08
bof
46.7k349113
1
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
4
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
1
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
2
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
 |Â
show 3 more comments
up vote
9
down vote
Your sequence $1,2,4,5,7,8,dots$ satisfies the second order nonhomogeneous linear recurrence
$$a_n=a_n-2+3$$
with initial values
$$a_1=1, a_2=2.$$
The solution is
$$a_n=frac6n-3-(-1)^n4.$$
The sum of the first $n$ terms is
$$sum_k=1^n a_k=frac34n^2+frac1-(-1)^n8=boxedfrac34n^2+frac1-cos npi8.$$
answered Sep 2 at 4:08
bof
46.7k349113
1
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
4
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
1
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
2
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
 |Â
show 3 more comments
up vote
9
down vote
up vote
9
down vote
Your sequence $1,2,4,5,7,8,dots$ satisfies the second order nonhomogeneous linear recurrence
$$a_n=a_n-2+3$$
with initial values
$$a_1=1, a_2=2.$$
The solution is
$$a_n=frac6n-3-(-1)^n4.$$
The sum of the first $n$ terms is
$$sum_k=1^n a_k=frac34n^2+frac1-(-1)^n8=boxedfrac34n^2+frac1-cos npi8.$$
answered Sep 2 at 4:08
bof
46.7k349113
Your sequence $1,2,4,5,7,8,dots$ satisfies the second order nonhomogeneous linear recurrence
$$a_n=a_n-2+3$$
with initial values
$$a_1=1, a_2=2.$$
The solution is
$$a_n=frac6n-3-(-1)^n4.$$
The sum of the first $n$ terms is
$$sum_k=1^n a_k=frac34n^2+frac1-(-1)^n8=boxedfrac34n^2+frac1-cos npi8.$$
answered Sep 2 at 4:08
bof
46.7k349113
edited Sep 2 at 5:32
answered Sep 2 at 4:08
bof
46.7k349113
answered Sep 2 at 4:08
bof
46.7k349113
answered Sep 2 at 4:08
bof
46.7k349113
46.7k349113
1
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
4
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
1
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
2
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
 |Â
show 3 more comments
1
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
4
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
1
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
2
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
1
1
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Did you guess this sequence or can you conclude this formula for $a_n$ once you notice it satisfies this second order nonhomogeneous linear recurrence?
– Cornman
Sep 2 at 4:11
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
Thanks. I am not familiar with such homogeneous equations. I think this is some valuable inside!
– Cornman
Sep 2 at 4:17
4
4
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
@bof $(-1)^n$ is usually considered the simpler, nicer form, especially in combinatorics.
– mephistolotl
Sep 2 at 6:46
1
1
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
@bof Don't you think $(-1)^n$ is good enough?
– Szeto
Sep 2 at 6:46
2
2
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
@bof Using an extra $cos$ does not make your answer bad.
– Szeto
Sep 2 at 8:39
 |Â
show 3 more comments
up vote
4
down vote
If you consider
$$a_k=1+3kqquad b_k=2+3k$$ then
$$sum_k=0^n a_k=sum_k=0^n (1+3k)=sum_k=0^n 1+3sum_k=0^n k=n+1+3 frac12 n (n+1)=frac12 (n+1) (3 n+2)$$
$$sum_k=0^n b_k=sum_k=0^n (2+3k)=2sum_k=0^n 1+3sum_k=0^n k=2(n+1)+3 frac12 n (n+1)=frac12 (n+1) (3 n+4)$$
Just continue.
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
add a comment |Â
up vote
4
down vote
If you consider
$$a_k=1+3kqquad b_k=2+3k$$ then
$$sum_k=0^n a_k=sum_k=0^n (1+3k)=sum_k=0^n 1+3sum_k=0^n k=n+1+3 frac12 n (n+1)=frac12 (n+1) (3 n+2)$$
$$sum_k=0^n b_k=sum_k=0^n (2+3k)=2sum_k=0^n 1+3sum_k=0^n k=2(n+1)+3 frac12 n (n+1)=frac12 (n+1) (3 n+4)$$
Just continue.
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
add a comment |Â
up vote
4
down vote
up vote
4
down vote
If you consider
$$a_k=1+3kqquad b_k=2+3k$$ then
$$sum_k=0^n a_k=sum_k=0^n (1+3k)=sum_k=0^n 1+3sum_k=0^n k=n+1+3 frac12 n (n+1)=frac12 (n+1) (3 n+2)$$
$$sum_k=0^n b_k=sum_k=0^n (2+3k)=2sum_k=0^n 1+3sum_k=0^n k=2(n+1)+3 frac12 n (n+1)=frac12 (n+1) (3 n+4)$$
Just continue.
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
If you consider
$$a_k=1+3kqquad b_k=2+3k$$ then
$$sum_k=0^n a_k=sum_k=0^n (1+3k)=sum_k=0^n 1+3sum_k=0^n k=n+1+3 frac12 n (n+1)=frac12 (n+1) (3 n+2)$$
$$sum_k=0^n b_k=sum_k=0^n (2+3k)=2sum_k=0^n 1+3sum_k=0^n k=2(n+1)+3 frac12 n (n+1)=frac12 (n+1) (3 n+4)$$
Just continue.
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
answered Sep 2 at 4:16
Claude Leibovici
113k1155127
113k1155127
add a comment |Â
add a comment |Â
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9
You should define your sequence clearly. I assume it's just the naturals less the multiples of $3$ but why make your readers guess? If I am correct, the answer is then the difference between a triangular number and three times another triangular number.
– lulu
Sep 2 at 3:50