A sum with geometric progression [closed]
$begingroup$
Can anyone help me in computing this sum?
$$sum_{k=1}^{l}4^{k}3^{l+1-k}$$
abstract-algebra summation
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
$endgroup$
closed as off-topic by José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
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." – José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Can anyone help me in computing this sum?
$$sum_{k=1}^{l}4^{k}3^{l+1-k}$$
abstract-algebra summation
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
$endgroup$
closed as off-topic by José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
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." – José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
$3^{l+1}sum_{k=1}^l(frac{4}{3})^k$
$endgroup$
– W. mu
Jan 4 at 8:22
$begingroup$
Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression?
$endgroup$
– Theo Bendit
Jan 4 at 8:22
2
$begingroup$
Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.)
$endgroup$
– Theo Bendit
Jan 4 at 8:23
$begingroup$
Abstract algebra is a misleading tag
$endgroup$
– roman
Jan 4 at 8:24
add a comment |
$begingroup$
Can anyone help me in computing this sum?
$$sum_{k=1}^{l}4^{k}3^{l+1-k}$$
abstract-algebra summation
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
$endgroup$
Can anyone help me in computing this sum?
$$sum_{k=1}^{l}4^{k}3^{l+1-k}$$
abstract-algebra summation
abstract-algebra summation
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
edited Jan 4 at 9:23
Le Anh Dung
1,4511621
1,4511621
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
asked Jan 4 at 8:20
Găitan NicolaiGăitan Nicolai
11
11
closed as off-topic by José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
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." – José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
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." – José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
$3^{l+1}sum_{k=1}^l(frac{4}{3})^k$
$endgroup$
– W. mu
Jan 4 at 8:22
$begingroup$
Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression?
$endgroup$
– Theo Bendit
Jan 4 at 8:22
2
$begingroup$
Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.)
$endgroup$
– Theo Bendit
Jan 4 at 8:23
$begingroup$
Abstract algebra is a misleading tag
$endgroup$
– roman
Jan 4 at 8:24
add a comment |
4
$begingroup$
$3^{l+1}sum_{k=1}^l(frac{4}{3})^k$
$endgroup$
– W. mu
Jan 4 at 8:22
$begingroup$
Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression?
$endgroup$
– Theo Bendit
Jan 4 at 8:22
2
$begingroup$
Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.)
$endgroup$
– Theo Bendit
Jan 4 at 8:23
$begingroup$
Abstract algebra is a misleading tag
$endgroup$
– roman
Jan 4 at 8:24
4
4
$begingroup$
$3^{l+1}sum_{k=1}^l(frac{4}{3})^k$
$endgroup$
– W. mu
Jan 4 at 8:22
$begingroup$
$3^{l+1}sum_{k=1}^l(frac{4}{3})^k$
$endgroup$
– W. mu
Jan 4 at 8:22
$begingroup$
Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression?
$endgroup$
– Theo Bendit
Jan 4 at 8:22
$begingroup$
Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression?
$endgroup$
– Theo Bendit
Jan 4 at 8:22
2
2
$begingroup$
Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.)
$endgroup$
– Theo Bendit
Jan 4 at 8:23
$begingroup$
Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.)
$endgroup$
– Theo Bendit
Jan 4 at 8:23
$begingroup$
Abstract algebra is a misleading tag
$endgroup$
– roman
Jan 4 at 8:24
$begingroup$
Abstract algebra is a misleading tag
$endgroup$
– roman
Jan 4 at 8:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$
sum_{k=1}^n 4^k3^{n+1-k} = 3^{n+1}sum_{k=1}^n left({4over 3}right)^k \
= 3^{n+1}left({4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^nright)
$$
Which is a regular geometric sum as you noticed:
$$
begin{align}
{4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^n
&= frac{{4over 3}left(1-left({4over 3}right)^nright)}{1-{4over 3}} \
&= 4left({4over 3}right)^n - 4
end{align}
$$
edited Jan 4 at 9:38
Saad
20.4k92352
answered Jan 4 at 8:26
romanroman
2,43721226
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
sum_{k=1}^n 4^k3^{n+1-k} = 3^{n+1}sum_{k=1}^n left({4over 3}right)^k \
= 3^{n+1}left({4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^nright)
$$
Which is a regular geometric sum as you noticed:
$$
begin{align}
{4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^n
&= frac{{4over 3}left(1-left({4over 3}right)^nright)}{1-{4over 3}} \
&= 4left({4over 3}right)^n - 4
end{align}
$$
edited Jan 4 at 9:38
Saad
20.4k92352
answered Jan 4 at 8:26
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=1}^n 4^k3^{n+1-k} = 3^{n+1}sum_{k=1}^n left({4over 3}right)^k \
= 3^{n+1}left({4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^nright)
$$
Which is a regular geometric sum as you noticed:
$$
begin{align}
{4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^n
&= frac{{4over 3}left(1-left({4over 3}right)^nright)}{1-{4over 3}} \
&= 4left({4over 3}right)^n - 4
end{align}
$$
edited Jan 4 at 9:38
Saad
20.4k92352
answered Jan 4 at 8:26
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=1}^n 4^k3^{n+1-k} = 3^{n+1}sum_{k=1}^n left({4over 3}right)^k \
= 3^{n+1}left({4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^nright)
$$
Which is a regular geometric sum as you noticed:
$$
begin{align}
{4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^n
&= frac{{4over 3}left(1-left({4over 3}right)^nright)}{1-{4over 3}} \
&= 4left({4over 3}right)^n - 4
end{align}
$$
edited Jan 4 at 9:38
Saad
20.4k92352
answered Jan 4 at 8:26
romanroman
2,43721226
$endgroup$
$$
sum_{k=1}^n 4^k3^{n+1-k} = 3^{n+1}sum_{k=1}^n left({4over 3}right)^k \
= 3^{n+1}left({4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^nright)
$$
Which is a regular geometric sum as you noticed:
$$
begin{align}
{4over 3} + left(4over 3right)^2 + cdots + left(4over 3right)^n
&= frac{{4over 3}left(1-left({4over 3}right)^nright)}{1-{4over 3}} \
&= 4left({4over 3}right)^n - 4
end{align}
$$
edited Jan 4 at 9:38
Saad
20.4k92352
answered Jan 4 at 8:26
romanroman
2,43721226
edited Jan 4 at 9:38
Saad
20.4k92352
edited Jan 4 at 9:38
Saad
20.4k92352
edited Jan 4 at 9:38
Saad
20.4k92352
20.4k92352
answered Jan 4 at 8:26
romanroman
2,43721226
answered Jan 4 at 8:26
romanroman
2,43721226
answered Jan 4 at 8:26
romanroman
2,43721226
2,43721226
add a comment |
add a comment |
4
$begingroup$
$3^{l+1}sum_{k=1}^l(frac{4}{3})^k$
$endgroup$
– W. mu
Jan 4 at 8:22
$begingroup$
Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression?
$endgroup$
– Theo Bendit
Jan 4 at 8:22
2
$begingroup$
Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.)
$endgroup$
– Theo Bendit
Jan 4 at 8:23
$begingroup$
Abstract algebra is a misleading tag
$endgroup$
– roman
Jan 4 at 8:24