finding a closed formula to an expression and prove it by inductionInduction to prove a simple formula for calculating the $n$-th derivative?closed form solution for summation of $log(i)$maclaurin series and induction over binomial theoremProve by Induction using Baseline and splitting into LHS & RHS?Improper integral $int_0^inftyfracx^nx^m+n+1 dx=fracn! (m-1)!(m+n)!.$Proving a partial sum with inductionShow that $pi =4-sum_n=1^infty frac(n)!(n-1)!(2n+1)!2^n+1$Induction Proof of Taylor Series FormulaProve $sum_i=1^n(-1)^i(i-1)/2$ is boundedFinding inverse using quadratic formula
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finding a closed formula to an expression and prove it by induction
Induction to prove a simple formula for calculating the $n$-th derivative?closed form solution for summation of $log(i)$maclaurin series and induction over binomial theoremProve by Induction using Baseline and splitting into LHS & RHS?Improper integral $int_0^inftyfracx^nx^m+n+1 dx=fracn! (m-1)!(m+n)!.$Proving a partial sum with inductionShow that $pi =4-sum_n=1^infty frac(n)!(n-1)!(2n+1)!2^n+1$Induction Proof of Taylor Series FormulaProve $sum_i=1^n(-1)^i(i-1)/2$ is boundedFinding inverse using quadratic formula
$begingroup$
Let 𝑎 > 0, find a closed formula (no sigmas) for
$$sum_i=1^log_a n a^i $$
i found this formula:
$$ fraca(n-1)a-1$$
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:
$$ fraca(n-1)a-1$$
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:
$$ fraca(n-1)a-1$$
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:
$$ fraca(n-1)a-1$$
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 $log_an$ 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 :
$$displaystylesum^log_an_i=1a^i = dfraca(n-1)n-1 Rightarrow sum^log_an_i=1a^i + a^log_aan= dfraca(n-1)n-1 + an =
dfraca(n-1)+a^2n - ann-1 = dfraca(an-1)n-1 = sum^log_an+1_i=1a^i$$
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
yesterday
add a comment |
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$begingroup$
The formula you provided is accurate only when $log_an$ 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 :
$$displaystylesum^log_an_i=1a^i = dfraca(n-1)n-1 Rightarrow sum^log_an_i=1a^i + a^log_aan= dfraca(n-1)n-1 + an =
dfraca(n-1)+a^2n - ann-1 = dfraca(an-1)n-1 = sum^log_an+1_i=1a^i$$
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
yesterday
add a comment |
$begingroup$
The formula you provided is accurate only when $log_an$ 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 :
$$displaystylesum^log_an_i=1a^i = dfraca(n-1)n-1 Rightarrow sum^log_an_i=1a^i + a^log_aan= dfraca(n-1)n-1 + an =
dfraca(n-1)+a^2n - ann-1 = dfraca(an-1)n-1 = sum^log_an+1_i=1a^i$$
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
yesterday
add a comment |
$begingroup$
The formula you provided is accurate only when $log_an$ 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 :
$$displaystylesum^log_an_i=1a^i = dfraca(n-1)n-1 Rightarrow sum^log_an_i=1a^i + a^log_aan= dfraca(n-1)n-1 + an =
dfraca(n-1)+a^2n - ann-1 = dfraca(an-1)n-1 = sum^log_an+1_i=1a^i$$
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 $log_an$ 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 :
$$displaystylesum^log_an_i=1a^i = dfraca(n-1)n-1 Rightarrow sum^log_an_i=1a^i + a^log_aan= dfraca(n-1)n-1 + an =
dfraca(n-1)+a^2n - ann-1 = dfraca(an-1)n-1 = sum^log_an+1_i=1a^i$$
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.
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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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
yesterday
add a comment |
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
yesterday
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
yesterday
$begingroup$
you really helped me, thank you!
$endgroup$
– Exposed
yesterday
add a comment |
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