Mixing
Finding the formula for the sequence: 7, 10, 16, 25, etc, in terms of $n$, given…
$begingroup$
In terms of $n$, find the formula for the sequence $a_n$: 7, 10, 16, 25; given that the difference between each adjacent element in the sequence creates an arithmetic sequence of its own.
I got the incorrect solution and I cannot trace my mistake:
We create a new sequence $b_n$ based on the given information:
$b_1 = 10 - 7 = 3$
$b_2 = 16 - 10 = 6$
$b_3 = 25-16 = 9$
$d_{b_n} = 3$
We know that:
$$a_n - a_{n-1} = b_{n-1}$$
If we were to add all possible elements:
$$A_n - A_{n-1} = B_{n-1}$$
We also know that $A_n - A_{n-1} = a_n$, so therefore $a_n = B_{n-1}$:
$$a_n = frac{[2b_1 + d(n-2)](n-1)}{2} = boxed{frac{3n(n-1)}{2}}$$
However, this is incorrect as it gives values inconsistent with the definition of the sequence.
The correct answer is $frac{3n^2-3n+14}{2}$.
Where have I gone wrong?
sequences-and-series
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
$endgroup$
add a comment |
$begingroup$
In terms of $n$, find the formula for the sequence $a_n$: 7, 10, 16, 25; given that the difference between each adjacent element in the sequence creates an arithmetic sequence of its own.
I got the incorrect solution and I cannot trace my mistake:
We create a new sequence $b_n$ based on the given information:
$b_1 = 10 - 7 = 3$
$b_2 = 16 - 10 = 6$
$b_3 = 25-16 = 9$
$d_{b_n} = 3$
We know that:
$$a_n - a_{n-1} = b_{n-1}$$
If we were to add all possible elements:
$$A_n - A_{n-1} = B_{n-1}$$
We also know that $A_n - A_{n-1} = a_n$, so therefore $a_n = B_{n-1}$:
$$a_n = frac{[2b_1 + d(n-2)](n-1)}{2} = boxed{frac{3n(n-1)}{2}}$$
However, this is incorrect as it gives values inconsistent with the definition of the sequence.
The correct answer is $frac{3n^2-3n+14}{2}$.
Where have I gone wrong?
sequences-and-series
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
$endgroup$
$begingroup$
I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value.
$endgroup$
– lulu
Jan 11 at 17:05
$begingroup$
@lulu Isn't it implied from $b_1 = 10 - 7 = 3$?
$endgroup$
– daedsidog
Jan 11 at 17:07
$begingroup$
No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are.
$endgroup$
– lulu
Jan 11 at 17:17
add a comment |
$begingroup$
In terms of $n$, find the formula for the sequence $a_n$: 7, 10, 16, 25; given that the difference between each adjacent element in the sequence creates an arithmetic sequence of its own.
I got the incorrect solution and I cannot trace my mistake:
We create a new sequence $b_n$ based on the given information:
$b_1 = 10 - 7 = 3$
$b_2 = 16 - 10 = 6$
$b_3 = 25-16 = 9$
$d_{b_n} = 3$
We know that:
$$a_n - a_{n-1} = b_{n-1}$$
If we were to add all possible elements:
$$A_n - A_{n-1} = B_{n-1}$$
We also know that $A_n - A_{n-1} = a_n$, so therefore $a_n = B_{n-1}$:
$$a_n = frac{[2b_1 + d(n-2)](n-1)}{2} = boxed{frac{3n(n-1)}{2}}$$
However, this is incorrect as it gives values inconsistent with the definition of the sequence.
The correct answer is $frac{3n^2-3n+14}{2}$.
Where have I gone wrong?
sequences-and-series
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
$endgroup$
In terms of $n$, find the formula for the sequence $a_n$: 7, 10, 16, 25; given that the difference between each adjacent element in the sequence creates an arithmetic sequence of its own.
I got the incorrect solution and I cannot trace my mistake:
We create a new sequence $b_n$ based on the given information:
$b_1 = 10 - 7 = 3$
$b_2 = 16 - 10 = 6$
$b_3 = 25-16 = 9$
$d_{b_n} = 3$
We know that:
$$a_n - a_{n-1} = b_{n-1}$$
If we were to add all possible elements:
$$A_n - A_{n-1} = B_{n-1}$$
We also know that $A_n - A_{n-1} = a_n$, so therefore $a_n = B_{n-1}$:
$$a_n = frac{[2b_1 + d(n-2)](n-1)}{2} = boxed{frac{3n(n-1)}{2}}$$
However, this is incorrect as it gives values inconsistent with the definition of the sequence.
The correct answer is $frac{3n^2-3n+14}{2}$.
Where have I gone wrong?
sequences-and-series
sequences-and-series
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
asked Jan 11 at 17:03
daedsidogdaedsidog
29017
29017
$begingroup$
I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value.
$endgroup$
– lulu
Jan 11 at 17:05
$begingroup$
@lulu Isn't it implied from $b_1 = 10 - 7 = 3$?
$endgroup$
– daedsidog
Jan 11 at 17:07
$begingroup$
No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are.
$endgroup$
– lulu
Jan 11 at 17:17
add a comment |
$begingroup$
I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value.
$endgroup$
– lulu
Jan 11 at 17:05
$begingroup$
@lulu Isn't it implied from $b_1 = 10 - 7 = 3$?
$endgroup$
– daedsidog
Jan 11 at 17:07
$begingroup$
No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are.
$endgroup$
– lulu
Jan 11 at 17:17
$begingroup$
I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value.
$endgroup$
– lulu
Jan 11 at 17:05
$begingroup$
I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value.
$endgroup$
– lulu
Jan 11 at 17:05
$begingroup$
@lulu Isn't it implied from $b_1 = 10 - 7 = 3$?
$endgroup$
– daedsidog
Jan 11 at 17:07
$begingroup$
@lulu Isn't it implied from $b_1 = 10 - 7 = 3$?
$endgroup$
– daedsidog
Jan 11 at 17:07
$begingroup$
No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are.
$endgroup$
– lulu
Jan 11 at 17:17
$begingroup$
No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are.
$endgroup$
– lulu
Jan 11 at 17:17
add a comment |
1 Answer
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$begingroup$
You have an error when adding the sequences up to $n$.
You have the following:
$$
a_2 - a_1 = b_1 \
a_3 - a_2 = b_2 \
dots \
a_n - a_{n - 1} = b_{n - 1}
$$
So, when you add these equalities, you get:
$$sum_2^n a_i + sum_1^{n-1}a_i = sum_1^{n-1}b_i$$
The second term here is indeed $A_{n-1}$, and the right hand side is $B_{n=1}$, but the first is $A_n - a_1$. Therefore you have $A_n - a_1 - A_{n-1} = B_{n-1}$. Now, using $A_n - A_{n-1} = a_n$, we have
$$a_n = B_{n-1} + a_1 = frac{3n(n-1)}{2} + 7$$
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have an error when adding the sequences up to $n$.
You have the following:
$$
a_2 - a_1 = b_1 \
a_3 - a_2 = b_2 \
dots \
a_n - a_{n - 1} = b_{n - 1}
$$
So, when you add these equalities, you get:
$$sum_2^n a_i + sum_1^{n-1}a_i = sum_1^{n-1}b_i$$
The second term here is indeed $A_{n-1}$, and the right hand side is $B_{n=1}$, but the first is $A_n - a_1$. Therefore you have $A_n - a_1 - A_{n-1} = B_{n-1}$. Now, using $A_n - A_{n-1} = a_n$, we have
$$a_n = B_{n-1} + a_1 = frac{3n(n-1)}{2} + 7$$
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
$endgroup$
add a comment |
$begingroup$
You have an error when adding the sequences up to $n$.
You have the following:
$$
a_2 - a_1 = b_1 \
a_3 - a_2 = b_2 \
dots \
a_n - a_{n - 1} = b_{n - 1}
$$
So, when you add these equalities, you get:
$$sum_2^n a_i + sum_1^{n-1}a_i = sum_1^{n-1}b_i$$
The second term here is indeed $A_{n-1}$, and the right hand side is $B_{n=1}$, but the first is $A_n - a_1$. Therefore you have $A_n - a_1 - A_{n-1} = B_{n-1}$. Now, using $A_n - A_{n-1} = a_n$, we have
$$a_n = B_{n-1} + a_1 = frac{3n(n-1)}{2} + 7$$
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
$endgroup$
add a comment |
$begingroup$
You have an error when adding the sequences up to $n$.
You have the following:
$$
a_2 - a_1 = b_1 \
a_3 - a_2 = b_2 \
dots \
a_n - a_{n - 1} = b_{n - 1}
$$
So, when you add these equalities, you get:
$$sum_2^n a_i + sum_1^{n-1}a_i = sum_1^{n-1}b_i$$
The second term here is indeed $A_{n-1}$, and the right hand side is $B_{n=1}$, but the first is $A_n - a_1$. Therefore you have $A_n - a_1 - A_{n-1} = B_{n-1}$. Now, using $A_n - A_{n-1} = a_n$, we have
$$a_n = B_{n-1} + a_1 = frac{3n(n-1)}{2} + 7$$
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
$endgroup$
You have an error when adding the sequences up to $n$.
You have the following:
$$
a_2 - a_1 = b_1 \
a_3 - a_2 = b_2 \
dots \
a_n - a_{n - 1} = b_{n - 1}
$$
So, when you add these equalities, you get:
$$sum_2^n a_i + sum_1^{n-1}a_i = sum_1^{n-1}b_i$$
The second term here is indeed $A_{n-1}$, and the right hand side is $B_{n=1}$, but the first is $A_n - a_1$. Therefore you have $A_n - a_1 - A_{n-1} = B_{n-1}$. Now, using $A_n - A_{n-1} = a_n$, we have
$$a_n = B_{n-1} + a_1 = frac{3n(n-1)}{2} + 7$$
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
answered Jan 11 at 17:12
Todor MarkovTodor Markov
2,201411
2,201411
add a comment |
add a comment |
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$begingroup$
I don't see where you used the initial condition $a_1=7$. Just knowing the differences between terms in a sequence does not determine the sequence uniquely. You still need some more information, like a starting value.
$endgroup$
– lulu
Jan 11 at 17:05
$begingroup$
@lulu Isn't it implied from $b_1 = 10 - 7 = 3$?
$endgroup$
– daedsidog
Jan 11 at 17:07
$begingroup$
No. Just knowing that $a_2-a_1=3$ doesn't tell you what $a_1,a_2$ are.
$endgroup$
– lulu
Jan 11 at 17:17