Mixing
Finding the difference $(a_2 + a_4 + … + a_{100}) - (a_1 + a_3 + … + a_{99})$
$begingroup$
Given the general sequence $a_n = 4n^2 +6n - 4$, calculate the difference $(a_2 + a_4 + ... + a_{100}) - (a_1 + a_3 + ... + a_{99})$.
If we look closely the difference can be rewritten as the sum of the differences between every element:
$(a_2 - a_1) + (a_4 - a_3) + ... +(a_{100} - a_{99})$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
$$c_n = a_{n+1} - a_n = [4(n+1)^2 +6(n+1) - 4] -[4n^2 +6n - 4] = 8n + 10$$
The new sequence is an arithmetic one, where $c_1 = 18$, $d = 8$, and has 50 elements.
Using the summation formula for an arithmetic sequence, we get:
$$C_n = frac{c_1+c_{n}}{2}cdot nrightarrow C_{50} = frac{c_1+c_{50}}{2}cdot 50 = frac{18 + 410}{2}cdot 50 = boxed{10700}$$
This appears to be wrong however, as the correct answer says 20500.
Where have I gone wrong?
sequences-and-series
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
$endgroup$
add a comment |
$begingroup$
Given the general sequence $a_n = 4n^2 +6n - 4$, calculate the difference $(a_2 + a_4 + ... + a_{100}) - (a_1 + a_3 + ... + a_{99})$.
If we look closely the difference can be rewritten as the sum of the differences between every element:
$(a_2 - a_1) + (a_4 - a_3) + ... +(a_{100} - a_{99})$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
$$c_n = a_{n+1} - a_n = [4(n+1)^2 +6(n+1) - 4] -[4n^2 +6n - 4] = 8n + 10$$
The new sequence is an arithmetic one, where $c_1 = 18$, $d = 8$, and has 50 elements.
Using the summation formula for an arithmetic sequence, we get:
$$C_n = frac{c_1+c_{n}}{2}cdot nrightarrow C_{50} = frac{c_1+c_{50}}{2}cdot 50 = frac{18 + 410}{2}cdot 50 = boxed{10700}$$
This appears to be wrong however, as the correct answer says 20500.
Where have I gone wrong?
sequences-and-series
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
$endgroup$
2
$begingroup$
$6(n+1)neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more.
$endgroup$
– Dietrich Burde
Jan 21 at 19:12
$begingroup$
@DietrichBurde Thanks, but this is just a typographical error, my notes were different.
$endgroup$
– daedsidog
Jan 21 at 19:14
1
$begingroup$
Last element is not $c_{50}$ !
$endgroup$
– Damien
Jan 21 at 19:19
add a comment |
$begingroup$
Given the general sequence $a_n = 4n^2 +6n - 4$, calculate the difference $(a_2 + a_4 + ... + a_{100}) - (a_1 + a_3 + ... + a_{99})$.
If we look closely the difference can be rewritten as the sum of the differences between every element:
$(a_2 - a_1) + (a_4 - a_3) + ... +(a_{100} - a_{99})$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
$$c_n = a_{n+1} - a_n = [4(n+1)^2 +6(n+1) - 4] -[4n^2 +6n - 4] = 8n + 10$$
The new sequence is an arithmetic one, where $c_1 = 18$, $d = 8$, and has 50 elements.
Using the summation formula for an arithmetic sequence, we get:
$$C_n = frac{c_1+c_{n}}{2}cdot nrightarrow C_{50} = frac{c_1+c_{50}}{2}cdot 50 = frac{18 + 410}{2}cdot 50 = boxed{10700}$$
This appears to be wrong however, as the correct answer says 20500.
Where have I gone wrong?
sequences-and-series
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
$endgroup$
Given the general sequence $a_n = 4n^2 +6n - 4$, calculate the difference $(a_2 + a_4 + ... + a_{100}) - (a_1 + a_3 + ... + a_{99})$.
If we look closely the difference can be rewritten as the sum of the differences between every element:
$(a_2 - a_1) + (a_4 - a_3) + ... +(a_{100} - a_{99})$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
$$c_n = a_{n+1} - a_n = [4(n+1)^2 +6(n+1) - 4] -[4n^2 +6n - 4] = 8n + 10$$
The new sequence is an arithmetic one, where $c_1 = 18$, $d = 8$, and has 50 elements.
Using the summation formula for an arithmetic sequence, we get:
$$C_n = frac{c_1+c_{n}}{2}cdot nrightarrow C_{50} = frac{c_1+c_{50}}{2}cdot 50 = frac{18 + 410}{2}cdot 50 = boxed{10700}$$
This appears to be wrong however, as the correct answer says 20500.
Where have I gone wrong?
sequences-and-series
sequences-and-series
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
edited Jan 21 at 19:13
daedsidog
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
asked Jan 21 at 19:11
daedsidogdaedsidog
29517
29517
2
$begingroup$
$6(n+1)neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more.
$endgroup$
– Dietrich Burde
Jan 21 at 19:12
$begingroup$
@DietrichBurde Thanks, but this is just a typographical error, my notes were different.
$endgroup$
– daedsidog
Jan 21 at 19:14
1
$begingroup$
Last element is not $c_{50}$ !
$endgroup$
– Damien
Jan 21 at 19:19
add a comment |
2
$begingroup$
$6(n+1)neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more.
$endgroup$
– Dietrich Burde
Jan 21 at 19:12
$begingroup$
@DietrichBurde Thanks, but this is just a typographical error, my notes were different.
$endgroup$
– daedsidog
Jan 21 at 19:14
1
$begingroup$
Last element is not $c_{50}$ !
$endgroup$
– Damien
Jan 21 at 19:19
2
2
$begingroup$
$6(n+1)neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more.
$endgroup$
– Dietrich Burde
Jan 21 at 19:12
$begingroup$
$6(n+1)neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more.
$endgroup$
– Dietrich Burde
Jan 21 at 19:12
$begingroup$
@DietrichBurde Thanks, but this is just a typographical error, my notes were different.
$endgroup$
– daedsidog
Jan 21 at 19:14
$begingroup$
@DietrichBurde Thanks, but this is just a typographical error, my notes were different.
$endgroup$
– daedsidog
Jan 21 at 19:14
1
1
$begingroup$
Last element is not $c_{50}$ !
$endgroup$
– Damien
Jan 21 at 19:19
$begingroup$
Last element is not $c_{50}$ !
$endgroup$
– Damien
Jan 21 at 19:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
Herein lies the problem. What you ultimately end up doing is finding $c_1 + ... + c_{50}$, but notice:
$$c_1 + c_2 = (a_2 - a_1) + (a_3 - a_2) = a_3 - a_1$$
Another example:
$$c_1 + ... + c_5 = (a_2 - a_1) + (a_3 - a_2) + ... + (a_6 - a_5) = a_6 - a_1$$
Basically, the sum over $c_n$ ends up being a telescoping sum of sorts, and you can see by now
$$c_1 + ... + c_{50} = a_{51} - a_1$$
which is not the sum you wanted to compute. Indeed, plugging in the appropriate values for $n$ shows $a_{51} - a_1 = 10,700$, your erroneous result.
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
$endgroup$
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
add a comment |
$begingroup$
$$sum_{r=1}^{50}(a_{2r}-a_{2r-1})=sum (4(4r-1)+6)=?$$
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
Herein lies the problem. What you ultimately end up doing is finding $c_1 + ... + c_{50}$, but notice:
$$c_1 + c_2 = (a_2 - a_1) + (a_3 - a_2) = a_3 - a_1$$
Another example:
$$c_1 + ... + c_5 = (a_2 - a_1) + (a_3 - a_2) + ... + (a_6 - a_5) = a_6 - a_1$$
Basically, the sum over $c_n$ ends up being a telescoping sum of sorts, and you can see by now
$$c_1 + ... + c_{50} = a_{51} - a_1$$
which is not the sum you wanted to compute. Indeed, plugging in the appropriate values for $n$ shows $a_{51} - a_1 = 10,700$, your erroneous result.
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
$endgroup$
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
add a comment |
$begingroup$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
Herein lies the problem. What you ultimately end up doing is finding $c_1 + ... + c_{50}$, but notice:
$$c_1 + c_2 = (a_2 - a_1) + (a_3 - a_2) = a_3 - a_1$$
Another example:
$$c_1 + ... + c_5 = (a_2 - a_1) + (a_3 - a_2) + ... + (a_6 - a_5) = a_6 - a_1$$
Basically, the sum over $c_n$ ends up being a telescoping sum of sorts, and you can see by now
$$c_1 + ... + c_{50} = a_{51} - a_1$$
which is not the sum you wanted to compute. Indeed, plugging in the appropriate values for $n$ shows $a_{51} - a_1 = 10,700$, your erroneous result.
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
$endgroup$
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
add a comment |
$begingroup$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
Herein lies the problem. What you ultimately end up doing is finding $c_1 + ... + c_{50}$, but notice:
$$c_1 + c_2 = (a_2 - a_1) + (a_3 - a_2) = a_3 - a_1$$
Another example:
$$c_1 + ... + c_5 = (a_2 - a_1) + (a_3 - a_2) + ... + (a_6 - a_5) = a_6 - a_1$$
Basically, the sum over $c_n$ ends up being a telescoping sum of sorts, and you can see by now
$$c_1 + ... + c_{50} = a_{51} - a_1$$
which is not the sum you wanted to compute. Indeed, plugging in the appropriate values for $n$ shows $a_{51} - a_1 = 10,700$, your erroneous result.
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
$endgroup$
We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:
Herein lies the problem. What you ultimately end up doing is finding $c_1 + ... + c_{50}$, but notice:
$$c_1 + c_2 = (a_2 - a_1) + (a_3 - a_2) = a_3 - a_1$$
Another example:
$$c_1 + ... + c_5 = (a_2 - a_1) + (a_3 - a_2) + ... + (a_6 - a_5) = a_6 - a_1$$
Basically, the sum over $c_n$ ends up being a telescoping sum of sorts, and you can see by now
$$c_1 + ... + c_{50} = a_{51} - a_1$$
which is not the sum you wanted to compute. Indeed, plugging in the appropriate values for $n$ shows $a_{51} - a_1 = 10,700$, your erroneous result.
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
answered Jan 21 at 19:19
Eevee TrainerEevee Trainer
7,59721338
7,59721338
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
add a comment |
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
$begingroup$
Ahh. Thank you.
$endgroup$
– daedsidog
Jan 21 at 19:24
add a comment |
$begingroup$
$$sum_{r=1}^{50}(a_{2r}-a_{2r-1})=sum (4(4r-1)+6)=?$$
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
$$sum_{r=1}^{50}(a_{2r}-a_{2r-1})=sum (4(4r-1)+6)=?$$
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
$$sum_{r=1}^{50}(a_{2r}-a_{2r-1})=sum (4(4r-1)+6)=?$$
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$$sum_{r=1}^{50}(a_{2r}-a_{2r-1})=sum (4(4r-1)+6)=?$$
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 21 at 19:17
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
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2
$begingroup$
$6(n+1)neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more.
$endgroup$
– Dietrich Burde
Jan 21 at 19:12
$begingroup$
@DietrichBurde Thanks, but this is just a typographical error, my notes were different.
$endgroup$
– daedsidog
Jan 21 at 19:14
1
$begingroup$
Last element is not $c_{50}$ !
$endgroup$
– Damien
Jan 21 at 19:19