How to write a general formula for $-2/9, 3/16, -4/25, 5/36…$? [on hold]Proving that for every real $x$ there exists $y$ with $x+y^2inmathbbQ$Reversing an Arithmetic SequenceFinding a formula for a given series.Quartic Solution on Wikipedia special cases problem $S=0$ how to “change the choice of cubic root”?Sequences: how to prove it is crescent and calculate the limitLimit $(n - a_n)$ of sequence $a_n+1 = sqrtn^2 - a_n$Finding the general termAlmost Alternating sequence general formulaUse Ratio test to find convergence of $sum_k=1^infty frac3x^kk^3$Formula for the sequence 0,3,8,15,24 …
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How to write a general formula for $-2/9, 3/16, -4/25, 5/36…$? [on hold]
Proving that for every real $x$ there exists $y$ with $x+y^2inmathbbQ$Reversing an Arithmetic SequenceFinding a formula for a given series.Quartic Solution on Wikipedia special cases problem $S=0$ how to “change the choice of cubic root”?Sequences: how to prove it is crescent and calculate the limitLimit $(n - a_n)$ of sequence $a_n+1 = sqrtn^2 - a_n$Finding the general termAlmost Alternating sequence general formulaUse Ratio test to find convergence of $sum_k=1^infty frac3x^kk^3$Formula for the sequence 0,3,8,15,24 …
$begingroup$
I'm having a bit of trouble with this sequence:
$-2/9, 3/16, -4/25, 5/36...$
I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change
sequences-and-series algebra-precalculus
$endgroup$
put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel
add a comment |
$begingroup$
I'm having a bit of trouble with this sequence:
$-2/9, 3/16, -4/25, 5/36...$
I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change
sequences-and-series algebra-precalculus
$endgroup$
put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel
3
$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
yesterday
$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
yesterday
add a comment |
$begingroup$
I'm having a bit of trouble with this sequence:
$-2/9, 3/16, -4/25, 5/36...$
I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change
sequences-and-series algebra-precalculus
$endgroup$
I'm having a bit of trouble with this sequence:
$-2/9, 3/16, -4/25, 5/36...$
I've gotten: $(n+1)/(n+2)^2$, but I'm not sure how to account for the sign change
sequences-and-series algebra-precalculus
sequences-and-series algebra-precalculus
edited yesterday
user21820
39.4k543155
39.4k543155
asked yesterday
vmahajan17vmahajan17
218
218
put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel
put on hold as off-topic by Saad, RRL, user21820, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, user21820, José Carlos Santos, Parcly Taxel
3
$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
yesterday
$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
yesterday
add a comment |
3
$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
yesterday
$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
yesterday
3
3
$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
yesterday
$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
yesterday
$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
yesterday
$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
yesterday
add a comment |
3 Answers
3
active
oldest
votes
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Answer is $$(-1)^nfracn+1(n+2)^2$$
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add a comment |
$begingroup$
Here is the way:
$$
a_n=(-1)^nfracn+1(n+2)^2
$$
$endgroup$
add a comment |
$begingroup$
$$T_n=(-1)^nfracn+1(n+2^2)$$
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Answer is $$(-1)^nfracn+1(n+2)^2$$
$endgroup$
add a comment |
$begingroup$
Answer is $$(-1)^nfracn+1(n+2)^2$$
$endgroup$
add a comment |
$begingroup$
Answer is $$(-1)^nfracn+1(n+2)^2$$
$endgroup$
Answer is $$(-1)^nfracn+1(n+2)^2$$
answered yesterday
Sujit BhattacharyyaSujit Bhattacharyya
1,533519
1,533519
add a comment |
add a comment |
$begingroup$
Here is the way:
$$
a_n=(-1)^nfracn+1(n+2)^2
$$
$endgroup$
add a comment |
$begingroup$
Here is the way:
$$
a_n=(-1)^nfracn+1(n+2)^2
$$
$endgroup$
add a comment |
$begingroup$
Here is the way:
$$
a_n=(-1)^nfracn+1(n+2)^2
$$
$endgroup$
Here is the way:
$$
a_n=(-1)^nfracn+1(n+2)^2
$$
answered yesterday
Holding ArthurHolding Arthur
1,271417
1,271417
add a comment |
add a comment |
$begingroup$
$$T_n=(-1)^nfracn+1(n+2^2)$$
$endgroup$
add a comment |
$begingroup$
$$T_n=(-1)^nfracn+1(n+2^2)$$
$endgroup$
add a comment |
$begingroup$
$$T_n=(-1)^nfracn+1(n+2^2)$$
$endgroup$
$$T_n=(-1)^nfracn+1(n+2^2)$$
answered yesterday
saket kumarsaket kumar
198113
198113
add a comment |
add a comment |
3
$begingroup$
How about $(-1)^n$?
$endgroup$
– Pebeto
yesterday
$begingroup$
ah makes sense. thanks!
$endgroup$
– vmahajan17
yesterday