finding a closed formula to an expression and prove it by inductionInduction to prove a simple formula for...
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finding a closed formula to an expression and prove it by induction
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$begingroup$
Let 𝑎 > 0, find a closed formula (no sigmas) for
$$sum_{i=1}^{log_a n} a^i $$
i found this formula:
$$ frac{a(n-1)}{a-1}$$
but i have trouble proving by induction
i dont really understand what is the base case, the hypothesis and the step
please help, thanks
calculus
asked Mar 11 at 19:18
ExposedExposed
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let 𝑎 > 0, find a closed formula (no sigmas) for
$$sum_{i=1}^{log_a n} a^i $$
i found this formula:
$$ frac{a(n-1)}{a-1}$$
but i have trouble proving by induction
i dont really understand what is the base case, the hypothesis and the step
please help, thanks
calculus
asked Mar 11 at 19:18
ExposedExposed
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Use formula for sum of first k terms of geometric progression where k=log_a(n)
$endgroup$
– Vladislav
Mar 11 at 19:24
$begingroup$
What is exactly the upper summation limit?
$endgroup$
– user
Mar 11 at 19:38
add a comment |
$begingroup$
Let 𝑎 > 0, find a closed formula (no sigmas) for
$$sum_{i=1}^{log_a n} a^i $$
i found this formula:
$$ frac{a(n-1)}{a-1}$$
but i have trouble proving by induction
i dont really understand what is the base case, the hypothesis and the step
please help, thanks
calculus
asked Mar 11 at 19:18
ExposedExposed
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let 𝑎 > 0, find a closed formula (no sigmas) for
$$sum_{i=1}^{log_a n} a^i $$
i found this formula:
$$ frac{a(n-1)}{a-1}$$
but i have trouble proving by induction
i dont really understand what is the base case, the hypothesis and the step
please help, thanks
calculus
calculus
asked Mar 11 at 19:18
ExposedExposed
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 19:18
ExposedExposed
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 19:18
ExposedExposed
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 11 at 19:18
ExposedExposed
1
asked Mar 11 at 19:18
ExposedExposed
1
1
New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Exposed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$begingroup$
Use formula for sum of first k terms of geometric progression where k=log_a(n)
$endgroup$
– Vladislav
Mar 11 at 19:24
$begingroup$
What is exactly the upper summation limit?
$endgroup$
– user
Mar 11 at 19:38
add a comment |
1
$begingroup$
Use formula for sum of first k terms of geometric progression where k=log_a(n)
$endgroup$
– Vladislav
Mar 11 at 19:24
$begingroup$
What is exactly the upper summation limit?
$endgroup$
– user
Mar 11 at 19:38
1
1
$begingroup$
Use formula for sum of first k terms of geometric progression where k=log_a(n)
$endgroup$
– Vladislav
Mar 11 at 19:24
$begingroup$
Use formula for sum of first k terms of geometric progression where k=log_a(n)
$endgroup$
– Vladislav
Mar 11 at 19:24
$begingroup$
What is exactly the upper summation limit?
$endgroup$
– user
Mar 11 at 19:38
$begingroup$
What is exactly the upper summation limit?
$endgroup$
– user
Mar 11 at 19:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The formula you provided is accurate only when $log_a{n}$ is an integer. So, to solve it by induction, you will need to prove that if the formula is true for $n$, it must also be true for $an$. For base case just using $n=1$ works.
Proof of Inductive Step :
$$displaystylesum^{log_a{n}}_{i=1}a^i = dfrac{a(n-1)}{n-1} Rightarrow sum^{log_a{n}}_{i=1}a^i + a^{log_a{an}}= dfrac{a(n-1)}{n-1} + an =
dfrac{a(n-1)+a^2n - an}{n-1} = dfrac{a(an-1)}{n-1} = sum^{log_a{n}+1}_{i=1}a^i$$
answered Mar 11 at 19:34
Siam HabibSiam Habib
113
New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
add a comment |
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1 Answer
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$begingroup$
The formula you provided is accurate only when $log_a{n}$ is an integer. So, to solve it by induction, you will need to prove that if the formula is true for $n$, it must also be true for $an$. For base case just using $n=1$ works.
Proof of Inductive Step :
$$displaystylesum^{log_a{n}}_{i=1}a^i = dfrac{a(n-1)}{n-1} Rightarrow sum^{log_a{n}}_{i=1}a^i + a^{log_a{an}}= dfrac{a(n-1)}{n-1} + an =
dfrac{a(n-1)+a^2n - an}{n-1} = dfrac{a(an-1)}{n-1} = sum^{log_a{n}+1}_{i=1}a^i$$
answered Mar 11 at 19:34
Siam HabibSiam Habib
113
New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
add a comment |
$begingroup$
The formula you provided is accurate only when $log_a{n}$ is an integer. So, to solve it by induction, you will need to prove that if the formula is true for $n$, it must also be true for $an$. For base case just using $n=1$ works.
Proof of Inductive Step :
$$displaystylesum^{log_a{n}}_{i=1}a^i = dfrac{a(n-1)}{n-1} Rightarrow sum^{log_a{n}}_{i=1}a^i + a^{log_a{an}}= dfrac{a(n-1)}{n-1} + an =
dfrac{a(n-1)+a^2n - an}{n-1} = dfrac{a(an-1)}{n-1} = sum^{log_a{n}+1}_{i=1}a^i$$
answered Mar 11 at 19:34
Siam HabibSiam Habib
113
New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
add a comment |
$begingroup$
The formula you provided is accurate only when $log_a{n}$ is an integer. So, to solve it by induction, you will need to prove that if the formula is true for $n$, it must also be true for $an$. For base case just using $n=1$ works.
Proof of Inductive Step :
$$displaystylesum^{log_a{n}}_{i=1}a^i = dfrac{a(n-1)}{n-1} Rightarrow sum^{log_a{n}}_{i=1}a^i + a^{log_a{an}}= dfrac{a(n-1)}{n-1} + an =
dfrac{a(n-1)+a^2n - an}{n-1} = dfrac{a(an-1)}{n-1} = sum^{log_a{n}+1}_{i=1}a^i$$
answered Mar 11 at 19:34
Siam HabibSiam Habib
113
New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The formula you provided is accurate only when $log_a{n}$ is an integer. So, to solve it by induction, you will need to prove that if the formula is true for $n$, it must also be true for $an$. For base case just using $n=1$ works.
Proof of Inductive Step :
$$displaystylesum^{log_a{n}}_{i=1}a^i = dfrac{a(n-1)}{n-1} Rightarrow sum^{log_a{n}}_{i=1}a^i + a^{log_a{an}}= dfrac{a(n-1)}{n-1} + an =
dfrac{a(n-1)+a^2n - an}{n-1} = dfrac{a(an-1)}{n-1} = sum^{log_a{n}+1}_{i=1}a^i$$
answered Mar 11 at 19:34
Siam HabibSiam Habib
113
New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Mar 11 at 19:34
Siam HabibSiam Habib
113
New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Mar 11 at 19:34
Siam HabibSiam Habib
113
answered Mar 11 at 19:34
Siam HabibSiam Habib
113
113
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Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Siam Habib is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
add a comment |
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
Mar 14 at 14:28
add a comment |
Exposed is a new contributor. Be nice, and check out our Code of Conduct.
Exposed is a new contributor. Be nice, and check out our Code of Conduct.
Exposed is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Use formula for sum of first k terms of geometric progression where k=log_a(n)
$endgroup$
– Vladislav
Mar 11 at 19:24
$begingroup$
What is exactly the upper summation limit?
$endgroup$
– user
Mar 11 at 19:38