Mixing
Method of differences summation
Clash Royale CLAN TAG#URR8PPP
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I was trying some Cambridge past papers and it said to first separate into partial fractions and then find the sum of the sequence, however after splitting inot partial fractions I'm not getting the terms to cancel out like I normally do with these questions. Is there something I'm missing
.been trying to manipulate the 3 sets of term but can't seem to get it . Thanks
$$sum_r=1^nfrac4r(r+1)(r+2)$$
sequences-and-series summation partial-fractions
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
asked Aug 4 at 2:29
user122343
445
 |Â
show 5 more comments
up vote
0
down vote
favorite
I was trying some Cambridge past papers and it said to first separate into partial fractions and then find the sum of the sequence, however after splitting inot partial fractions I'm not getting the terms to cancel out like I normally do with these questions. Is there something I'm missing
.been trying to manipulate the 3 sets of term but can't seem to get it . Thanks
$$sum_r=1^nfrac4r(r+1)(r+2)$$
sequences-and-series summation partial-fractions
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
asked Aug 4 at 2:29
user122343
445
What do you get after applying partial fractions?
– Simply Beautiful Art
Aug 4 at 2:36
(2/r) - 4/(r+1) + 2/r+2
– user122343
Aug 4 at 2:39
Can you figure out $sum a_r-a_r+1$? What about if $a_r=frac2r-frac2r+1$?
– Simply Beautiful Art
Aug 4 at 2:53
Yes I can do that since the terms cancel out easily
– user122343
Aug 4 at 2:56
Then you've figured out the answer? :-)
– Simply Beautiful Art
Aug 4 at 3:02
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was trying some Cambridge past papers and it said to first separate into partial fractions and then find the sum of the sequence, however after splitting inot partial fractions I'm not getting the terms to cancel out like I normally do with these questions. Is there something I'm missing
.been trying to manipulate the 3 sets of term but can't seem to get it . Thanks
$$sum_r=1^nfrac4r(r+1)(r+2)$$
sequences-and-series summation partial-fractions
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
asked Aug 4 at 2:29
user122343
445
I was trying some Cambridge past papers and it said to first separate into partial fractions and then find the sum of the sequence, however after splitting inot partial fractions I'm not getting the terms to cancel out like I normally do with these questions. Is there something I'm missing
.been trying to manipulate the 3 sets of term but can't seem to get it . Thanks
$$sum_r=1^nfrac4r(r+1)(r+2)$$
sequences-and-series summation partial-fractions
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
asked Aug 4 at 2:29
user122343
445
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
edited Aug 4 at 2:35
Simply Beautiful Art
49k571169
49k571169
asked Aug 4 at 2:29
user122343
445
asked Aug 4 at 2:29
user122343
445
asked Aug 4 at 2:29
user122343
445
445
What do you get after applying partial fractions?
– Simply Beautiful Art
Aug 4 at 2:36
(2/r) - 4/(r+1) + 2/r+2
– user122343
Aug 4 at 2:39
Can you figure out $sum a_r-a_r+1$? What about if $a_r=frac2r-frac2r+1$?
– Simply Beautiful Art
Aug 4 at 2:53
Yes I can do that since the terms cancel out easily
– user122343
Aug 4 at 2:56
Then you've figured out the answer? :-)
– Simply Beautiful Art
Aug 4 at 3:02
 |Â
show 5 more comments
What do you get after applying partial fractions?
– Simply Beautiful Art
Aug 4 at 2:36
(2/r) - 4/(r+1) + 2/r+2
– user122343
Aug 4 at 2:39
Can you figure out $sum a_r-a_r+1$? What about if $a_r=frac2r-frac2r+1$?
– Simply Beautiful Art
Aug 4 at 2:53
Yes I can do that since the terms cancel out easily
– user122343
Aug 4 at 2:56
Then you've figured out the answer? :-)
– Simply Beautiful Art
Aug 4 at 3:02
What do you get after applying partial fractions?
– Simply Beautiful Art
Aug 4 at 2:36
What do you get after applying partial fractions?
– Simply Beautiful Art
Aug 4 at 2:36
(2/r) - 4/(r+1) + 2/r+2
– user122343
Aug 4 at 2:39
(2/r) - 4/(r+1) + 2/r+2
– user122343
Aug 4 at 2:39
Can you figure out $sum a_r-a_r+1$? What about if $a_r=frac2r-frac2r+1$?
– Simply Beautiful Art
Aug 4 at 2:53
Can you figure out $sum a_r-a_r+1$? What about if $a_r=frac2r-frac2r+1$?
– Simply Beautiful Art
Aug 4 at 2:53
Yes I can do that since the terms cancel out easily
– user122343
Aug 4 at 2:56
Yes I can do that since the terms cancel out easily
– user122343
Aug 4 at 2:56
Then you've figured out the answer? :-)
– Simply Beautiful Art
Aug 4 at 3:02
Then you've figured out the answer? :-)
– Simply Beautiful Art
Aug 4 at 3:02
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Take
$$b_r=frac1r(r+1).$$
Then
$$b_r-b_r+1=frac1r(r+1)-frac1(r+1)(r+2)=frac2r(r+1)(r+2)=frac12a_r.$$
So
$$sum_r=1^na_r=sum_r=1^nfrac4r(r+1)(r+2)=
2sum_r=1^n left(frac1r(r+1)-frac1(r+1)(r+2)right)=
2left(frac12-frac1(n+1)(n+2)right).
$$
In fact, if you want to sum, via differences and telescopic summation,
any finite series with a general term of the form
$$frac1(r+k)(r+k+1)...(r+m-1)(r+m)$$
where $k,m$ arew integers with $k<m$, then take
$$b_r=frac1(r+k)(r+k+1)...(r+m-1)(r+m-1)$$
answered Aug 4 at 3:22
Medo
454112
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
add a comment |Â
up vote
0
down vote
Using calculus of finite differences with rising factorials,
Let $S_n = sum_r=1^nfrac4r(r+1)(r+2) = sum_r=1^n4r^-underline3$
Then by integrating, we get $S_n = 4fracr^underline-2-2 + C$
$S_n = C - frac2(n+2)(n+1)$
For $n = 1$, $S_n = 4/6 = C - 2/6 implies C = 1$
So $S_n = 1 - frac2(n+1)(n+2)$
answered 2 days ago
sku
988311
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Take
$$b_r=frac1r(r+1).$$
Then
$$b_r-b_r+1=frac1r(r+1)-frac1(r+1)(r+2)=frac2r(r+1)(r+2)=frac12a_r.$$
So
$$sum_r=1^na_r=sum_r=1^nfrac4r(r+1)(r+2)=
2sum_r=1^n left(frac1r(r+1)-frac1(r+1)(r+2)right)=
2left(frac12-frac1(n+1)(n+2)right).
$$
In fact, if you want to sum, via differences and telescopic summation,
any finite series with a general term of the form
$$frac1(r+k)(r+k+1)...(r+m-1)(r+m)$$
where $k,m$ arew integers with $k<m$, then take
$$b_r=frac1(r+k)(r+k+1)...(r+m-1)(r+m-1)$$
answered Aug 4 at 3:22
Medo
454112
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
add a comment |Â
up vote
0
down vote
accepted
Take
$$b_r=frac1r(r+1).$$
Then
$$b_r-b_r+1=frac1r(r+1)-frac1(r+1)(r+2)=frac2r(r+1)(r+2)=frac12a_r.$$
So
$$sum_r=1^na_r=sum_r=1^nfrac4r(r+1)(r+2)=
2sum_r=1^n left(frac1r(r+1)-frac1(r+1)(r+2)right)=
2left(frac12-frac1(n+1)(n+2)right).
$$
In fact, if you want to sum, via differences and telescopic summation,
any finite series with a general term of the form
$$frac1(r+k)(r+k+1)...(r+m-1)(r+m)$$
where $k,m$ arew integers with $k<m$, then take
$$b_r=frac1(r+k)(r+k+1)...(r+m-1)(r+m-1)$$
answered Aug 4 at 3:22
Medo
454112
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Take
$$b_r=frac1r(r+1).$$
Then
$$b_r-b_r+1=frac1r(r+1)-frac1(r+1)(r+2)=frac2r(r+1)(r+2)=frac12a_r.$$
So
$$sum_r=1^na_r=sum_r=1^nfrac4r(r+1)(r+2)=
2sum_r=1^n left(frac1r(r+1)-frac1(r+1)(r+2)right)=
2left(frac12-frac1(n+1)(n+2)right).
$$
In fact, if you want to sum, via differences and telescopic summation,
any finite series with a general term of the form
$$frac1(r+k)(r+k+1)...(r+m-1)(r+m)$$
where $k,m$ arew integers with $k<m$, then take
$$b_r=frac1(r+k)(r+k+1)...(r+m-1)(r+m-1)$$
answered Aug 4 at 3:22
Medo
454112
Take
$$b_r=frac1r(r+1).$$
Then
$$b_r-b_r+1=frac1r(r+1)-frac1(r+1)(r+2)=frac2r(r+1)(r+2)=frac12a_r.$$
So
$$sum_r=1^na_r=sum_r=1^nfrac4r(r+1)(r+2)=
2sum_r=1^n left(frac1r(r+1)-frac1(r+1)(r+2)right)=
2left(frac12-frac1(n+1)(n+2)right).
$$
In fact, if you want to sum, via differences and telescopic summation,
any finite series with a general term of the form
$$frac1(r+k)(r+k+1)...(r+m-1)(r+m)$$
where $k,m$ arew integers with $k<m$, then take
$$b_r=frac1(r+k)(r+k+1)...(r+m-1)(r+m-1)$$
answered Aug 4 at 3:22
Medo
454112
edited Aug 4 at 3:29
answered Aug 4 at 3:22
Medo
454112
answered Aug 4 at 3:22
Medo
454112
answered Aug 4 at 3:22
Medo
454112
454112
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
add a comment |Â
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
thanks alot I finally understand but is there any way of knowing what to let $a_n$ be ? Or how to bring these types of sequences into a form so that method of differences can apply ?
– user122343
Aug 4 at 3:25
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
@user122343 Mostly, yes there is, depending on the general term of course. It is better to do exercises and enjoy discovering the choice that works for the differences.
– Medo
Aug 4 at 3:33
add a comment |Â
up vote
0
down vote
Using calculus of finite differences with rising factorials,
Let $S_n = sum_r=1^nfrac4r(r+1)(r+2) = sum_r=1^n4r^-underline3$
Then by integrating, we get $S_n = 4fracr^underline-2-2 + C$
$S_n = C - frac2(n+2)(n+1)$
For $n = 1$, $S_n = 4/6 = C - 2/6 implies C = 1$
So $S_n = 1 - frac2(n+1)(n+2)$
answered 2 days ago
sku
988311
add a comment |Â
up vote
0
down vote
Using calculus of finite differences with rising factorials,
Let $S_n = sum_r=1^nfrac4r(r+1)(r+2) = sum_r=1^n4r^-underline3$
Then by integrating, we get $S_n = 4fracr^underline-2-2 + C$
$S_n = C - frac2(n+2)(n+1)$
For $n = 1$, $S_n = 4/6 = C - 2/6 implies C = 1$
So $S_n = 1 - frac2(n+1)(n+2)$
answered 2 days ago
sku
988311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using calculus of finite differences with rising factorials,
Let $S_n = sum_r=1^nfrac4r(r+1)(r+2) = sum_r=1^n4r^-underline3$
Then by integrating, we get $S_n = 4fracr^underline-2-2 + C$
$S_n = C - frac2(n+2)(n+1)$
For $n = 1$, $S_n = 4/6 = C - 2/6 implies C = 1$
So $S_n = 1 - frac2(n+1)(n+2)$
answered 2 days ago
sku
988311
Using calculus of finite differences with rising factorials,
Let $S_n = sum_r=1^nfrac4r(r+1)(r+2) = sum_r=1^n4r^-underline3$
Then by integrating, we get $S_n = 4fracr^underline-2-2 + C$
$S_n = C - frac2(n+2)(n+1)$
For $n = 1$, $S_n = 4/6 = C - 2/6 implies C = 1$
So $S_n = 1 - frac2(n+1)(n+2)$
answered 2 days ago
sku
988311
answered 2 days ago
sku
988311
answered 2 days ago
sku
988311
answered 2 days ago
sku
988311
988311
add a comment |Â
add a comment |Â
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What do you get after applying partial fractions?
– Simply Beautiful Art
Aug 4 at 2:36
(2/r) - 4/(r+1) + 2/r+2
– user122343
Aug 4 at 2:39
Can you figure out $sum a_r-a_r+1$? What about if $a_r=frac2r-frac2r+1$?
– Simply Beautiful Art
Aug 4 at 2:53
Yes I can do that since the terms cancel out easily
– user122343
Aug 4 at 2:56
Then you've figured out the answer? :-)
– Simply Beautiful Art
Aug 4 at 3:02