Sum of $50$ terms of $cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+…$ [duplicate]
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This question already has an answer here:
Explicitly finding the sum of $arctan(1/(n^2+n+1))$
2 answers
Find the sum of the first $50$ terms of the series $$cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....$$
$$
sum_1^{50}=cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....\
=tan^{-1}frac{1}{3}+tan^{-1}frac{1}{7}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}+.....=
$$
My reference gives the solution $tan^{-1}dfrac{5}{6}$, but I do not have any clue of doing it ?
Note: I know that $tan^{-1}x+tan^{-1}y=tan^{-1}dfrac{x+y}{1-xy}$ if $xy<1$.
sequences-and-series trigonometry inverse-function
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
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marked as duplicate by Lord Shark the Unknown, lab bhattacharjee sequences-and-series
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Dec 16 '18 at 5:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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This question already has an answer here:
Explicitly finding the sum of $arctan(1/(n^2+n+1))$
2 answers
Find the sum of the first $50$ terms of the series $$cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....$$
$$
sum_1^{50}=cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....\
=tan^{-1}frac{1}{3}+tan^{-1}frac{1}{7}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}+.....=
$$
My reference gives the solution $tan^{-1}dfrac{5}{6}$, but I do not have any clue of doing it ?
Note: I know that $tan^{-1}x+tan^{-1}y=tan^{-1}dfrac{x+y}{1-xy}$ if $xy<1$.
sequences-and-series trigonometry inverse-function
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
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marked as duplicate by Lord Shark the Unknown, lab bhattacharjee sequences-and-series
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Dec 16 '18 at 5:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Note that $$tan^{-1} x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
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– Qurultay
Dec 15 '18 at 5:33
1
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what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $sum$ notation?
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– clathratus
Dec 15 '18 at 5:36
1
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@clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$.
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– farruhota
Dec 15 '18 at 5:41
2
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$a_n=1+n+n^2$, I guess
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– Claude Leibovici
Dec 15 '18 at 5:45
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Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it.
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– Empy2
Dec 15 '18 at 5:51
add a comment |
$begingroup$
This question already has an answer here:
Explicitly finding the sum of $arctan(1/(n^2+n+1))$
2 answers
Find the sum of the first $50$ terms of the series $$cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....$$
$$
sum_1^{50}=cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....\
=tan^{-1}frac{1}{3}+tan^{-1}frac{1}{7}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}+.....=
$$
My reference gives the solution $tan^{-1}dfrac{5}{6}$, but I do not have any clue of doing it ?
Note: I know that $tan^{-1}x+tan^{-1}y=tan^{-1}dfrac{x+y}{1-xy}$ if $xy<1$.
sequences-and-series trigonometry inverse-function
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
$endgroup$
This question already has an answer here:
Explicitly finding the sum of $arctan(1/(n^2+n+1))$
2 answers
Find the sum of the first $50$ terms of the series $$cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....$$
$$
sum_1^{50}=cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21+.....\
=tan^{-1}frac{1}{3}+tan^{-1}frac{1}{7}+tan^{-1}frac{1}{13}+tan^{-1}frac{1}{21}+.....=
$$
My reference gives the solution $tan^{-1}dfrac{5}{6}$, but I do not have any clue of doing it ?
Note: I know that $tan^{-1}x+tan^{-1}y=tan^{-1}dfrac{x+y}{1-xy}$ if $xy<1$.
This question already has an answer here:
Explicitly finding the sum of $arctan(1/(n^2+n+1))$
2 answers
sequences-and-series trigonometry inverse-function
sequences-and-series trigonometry inverse-function
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
edited Dec 15 '18 at 5:45
ss1729
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
asked Dec 15 '18 at 5:24
ss1729ss1729
1,9601923
1,9601923
marked as duplicate by Lord Shark the Unknown, lab bhattacharjee sequences-and-series
Users with the sequences-and-series badge can single-handedly close sequences-and-series questions as duplicates and reopen them as needed.
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Dec 16 '18 at 5:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, lab bhattacharjee sequences-and-series
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Dec 16 '18 at 5:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Note that $$tan^{-1} x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
$endgroup$
– Qurultay
Dec 15 '18 at 5:33
1
$begingroup$
what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $sum$ notation?
$endgroup$
– clathratus
Dec 15 '18 at 5:36
1
$begingroup$
@clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$.
$endgroup$
– farruhota
Dec 15 '18 at 5:41
2
$begingroup$
$a_n=1+n+n^2$, I guess
$endgroup$
– Claude Leibovici
Dec 15 '18 at 5:45
$begingroup$
Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it.
$endgroup$
– Empy2
Dec 15 '18 at 5:51
add a comment |
1
$begingroup$
Note that $$tan^{-1} x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
$endgroup$
– Qurultay
Dec 15 '18 at 5:33
1
$begingroup$
what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $sum$ notation?
$endgroup$
– clathratus
Dec 15 '18 at 5:36
1
$begingroup$
@clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$.
$endgroup$
– farruhota
Dec 15 '18 at 5:41
2
$begingroup$
$a_n=1+n+n^2$, I guess
$endgroup$
– Claude Leibovici
Dec 15 '18 at 5:45
$begingroup$
Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it.
$endgroup$
– Empy2
Dec 15 '18 at 5:51
1
1
$begingroup$
Note that $$tan^{-1} x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
$endgroup$
– Qurultay
Dec 15 '18 at 5:33
$begingroup$
Note that $$tan^{-1} x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
$endgroup$
– Qurultay
Dec 15 '18 at 5:33
1
1
$begingroup$
what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $sum$ notation?
$endgroup$
– clathratus
Dec 15 '18 at 5:36
$begingroup$
what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $sum$ notation?
$endgroup$
– clathratus
Dec 15 '18 at 5:36
1
1
$begingroup$
@clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$.
$endgroup$
– farruhota
Dec 15 '18 at 5:41
$begingroup$
@clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$.
$endgroup$
– farruhota
Dec 15 '18 at 5:41
2
2
$begingroup$
$a_n=1+n+n^2$, I guess
$endgroup$
– Claude Leibovici
Dec 15 '18 at 5:45
$begingroup$
$a_n=1+n+n^2$, I guess
$endgroup$
– Claude Leibovici
Dec 15 '18 at 5:45
$begingroup$
Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it.
$endgroup$
– Empy2
Dec 15 '18 at 5:51
$begingroup$
Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it.
$endgroup$
– Empy2
Dec 15 '18 at 5:51
add a comment |
1 Answer
1
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oldest
votes
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Hint
Use the fact that $cot^{-1}(x)=tan^{-1}frac{1}{x}$
$tan^{-1}left(frac{1}{n^2+n+1}right) = tan^{-1}left(frac{(n+1)-n}{1+n(n+1)}right) =tan^{-1}(n+1) -tan^{-1}(n)$
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Use the fact that $cot^{-1}(x)=tan^{-1}frac{1}{x}$
$tan^{-1}left(frac{1}{n^2+n+1}right) = tan^{-1}left(frac{(n+1)-n}{1+n(n+1)}right) =tan^{-1}(n+1) -tan^{-1}(n)$
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
$endgroup$
add a comment |
$begingroup$
Hint
Use the fact that $cot^{-1}(x)=tan^{-1}frac{1}{x}$
$tan^{-1}left(frac{1}{n^2+n+1}right) = tan^{-1}left(frac{(n+1)-n}{1+n(n+1)}right) =tan^{-1}(n+1) -tan^{-1}(n)$
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
$endgroup$
add a comment |
$begingroup$
Hint
Use the fact that $cot^{-1}(x)=tan^{-1}frac{1}{x}$
$tan^{-1}left(frac{1}{n^2+n+1}right) = tan^{-1}left(frac{(n+1)-n}{1+n(n+1)}right) =tan^{-1}(n+1) -tan^{-1}(n)$
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
$endgroup$
Hint
Use the fact that $cot^{-1}(x)=tan^{-1}frac{1}{x}$
$tan^{-1}left(frac{1}{n^2+n+1}right) = tan^{-1}left(frac{(n+1)-n}{1+n(n+1)}right) =tan^{-1}(n+1) -tan^{-1}(n)$
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
answered Dec 15 '18 at 5:51
crskhrcrskhr
3,873925
3,873925
add a comment |
add a comment |
1
$begingroup$
Note that $$tan^{-1} x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
$endgroup$
– Qurultay
Dec 15 '18 at 5:33
1
$begingroup$
what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $sum$ notation?
$endgroup$
– clathratus
Dec 15 '18 at 5:36
1
$begingroup$
@clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$.
$endgroup$
– farruhota
Dec 15 '18 at 5:41
2
$begingroup$
$a_n=1+n+n^2$, I guess
$endgroup$
– Claude Leibovici
Dec 15 '18 at 5:45
$begingroup$
Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it.
$endgroup$
– Empy2
Dec 15 '18 at 5:51