Proving by induction that for all n in set N an<3 is trueproblem understanding induction proof for following recurrence sequence $fraca_n-1+a_n-22$Show that $a_n = 2^n + 3^n .$ Strong Induction for noobs!Prove the inequality for all natural numbers n using inductionProving an identity of Fibonacci Numbers by inductionProve by strong induction that $3^n$ divides $a_n$ for all integers $n ge 1$Strong Induction: Prove $a_n=(-3)^n$Proving using induction or strong induction on Fibonacci number propositionProve by induction that $n^3 leq 3^n$ for all integers $geq 3$Prove by induction that for all $ninBbb N$.Recursive induction for a sequence.
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Proving by induction that for all n in set N an
problem understanding induction proof for following recurrence sequence $fraca_n-1+a_n-22$Show that $a_n = 2^n + 3^n .$ Strong Induction for noobs!Prove the inequality for all natural numbers n using inductionProving an identity of Fibonacci Numbers by inductionProve by strong induction that $3^n$ divides $a_n$ for all integers $n ge 1$Strong Induction: Prove $a_n=(-3)^n$Proving using induction or strong induction on Fibonacci number propositionProve by induction that $n^3 leq 3^n$ for all integers $geq 3$Prove by induction that for all $ninBbb N$.Recursive induction for a sequence.
$begingroup$
Data :
$a_0 = 2$.
and for all n that in $N$ set $a_n+1 = sqrt3*a_n$.
prove that for every $n$ in $N$ $a_n<3$.
Now I know I need to use induction.
I did the first step and said that if $a_0=2$ and $2<3$ is really true, thats the base of my induction.
but I know I also need to use the series definition to prove the next step of the induction but im stuck there.
discrete-mathematics induction
edited Mar 28 at 19:52
Max
9871319
asked Mar 28 at 19:50
GilGil
103
$endgroup$
add a comment |
$begingroup$
Data :
$a_0 = 2$.
and for all n that in $N$ set $a_n+1 = sqrt3*a_n$.
prove that for every $n$ in $N$ $a_n<3$.
Now I know I need to use induction.
I did the first step and said that if $a_0=2$ and $2<3$ is really true, thats the base of my induction.
but I know I also need to use the series definition to prove the next step of the induction but im stuck there.
discrete-mathematics induction
edited Mar 28 at 19:52
Max
9871319
asked Mar 28 at 19:50
GilGil
103
$endgroup$
add a comment |
$begingroup$
Data :
$a_0 = 2$.
and for all n that in $N$ set $a_n+1 = sqrt3*a_n$.
prove that for every $n$ in $N$ $a_n<3$.
Now I know I need to use induction.
I did the first step and said that if $a_0=2$ and $2<3$ is really true, thats the base of my induction.
but I know I also need to use the series definition to prove the next step of the induction but im stuck there.
discrete-mathematics induction
edited Mar 28 at 19:52
Max
9871319
asked Mar 28 at 19:50
GilGil
103
$endgroup$
Data :
$a_0 = 2$.
and for all n that in $N$ set $a_n+1 = sqrt3*a_n$.
prove that for every $n$ in $N$ $a_n<3$.
Now I know I need to use induction.
I did the first step and said that if $a_0=2$ and $2<3$ is really true, thats the base of my induction.
but I know I also need to use the series definition to prove the next step of the induction but im stuck there.
discrete-mathematics induction
discrete-mathematics induction
edited Mar 28 at 19:52
Max
9871319
asked Mar 28 at 19:50
GilGil
103
edited Mar 28 at 19:52
Max
9871319
asked Mar 28 at 19:50
GilGil
103
edited Mar 28 at 19:52
Max
9871319
edited Mar 28 at 19:52
Max
9871319
edited Mar 28 at 19:52
Max
9871319
9871319
asked Mar 28 at 19:50
GilGil
103
asked Mar 28 at 19:50
GilGil
103
asked Mar 28 at 19:50
GilGil
103
103
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's assume the hypothesis is true for some $a_n$, i.e. $a_n < 3$. We need to verify it is also true for $a_n+1$.
But
$$
a_n+1 = sqrt3 a_n = sqrt3 cdot sqrta_n.
$$
Since $a_n < 3$ you have $sqrta_n < sqrt3$...
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
Here is the induction step spelled out.
Assuming $0<a_n<3$, show that $0<a_n+1=sqrt3a_n<3$
(I added the $0<$ just to make absolutely certain that the square root is allowed.)
answered Mar 28 at 20:00
ArthurArthur
121k7122208
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's assume the hypothesis is true for some $a_n$, i.e. $a_n < 3$. We need to verify it is also true for $a_n+1$.
But
$$
a_n+1 = sqrt3 a_n = sqrt3 cdot sqrta_n.
$$
Since $a_n < 3$ you have $sqrta_n < sqrt3$...
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
Let's assume the hypothesis is true for some $a_n$, i.e. $a_n < 3$. We need to verify it is also true for $a_n+1$.
But
$$
a_n+1 = sqrt3 a_n = sqrt3 cdot sqrta_n.
$$
Since $a_n < 3$ you have $sqrta_n < sqrt3$...
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
Let's assume the hypothesis is true for some $a_n$, i.e. $a_n < 3$. We need to verify it is also true for $a_n+1$.
But
$$
a_n+1 = sqrt3 a_n = sqrt3 cdot sqrta_n.
$$
Since $a_n < 3$ you have $sqrta_n < sqrt3$...
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
$endgroup$
Let's assume the hypothesis is true for some $a_n$, i.e. $a_n < 3$. We need to verify it is also true for $a_n+1$.
But
$$
a_n+1 = sqrt3 a_n = sqrt3 cdot sqrta_n.
$$
Since $a_n < 3$ you have $sqrta_n < sqrt3$...
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
answered Mar 28 at 20:00
gt6989bgt6989b
35.2k22557
35.2k22557
add a comment |
add a comment |
$begingroup$
Here is the induction step spelled out.
Assuming $0<a_n<3$, show that $0<a_n+1=sqrt3a_n<3$
(I added the $0<$ just to make absolutely certain that the square root is allowed.)
answered Mar 28 at 20:00
ArthurArthur
121k7122208
$endgroup$
add a comment |
$begingroup$
Here is the induction step spelled out.
Assuming $0<a_n<3$, show that $0<a_n+1=sqrt3a_n<3$
(I added the $0<$ just to make absolutely certain that the square root is allowed.)
answered Mar 28 at 20:00
ArthurArthur
121k7122208
$endgroup$
add a comment |
$begingroup$
Here is the induction step spelled out.
Assuming $0<a_n<3$, show that $0<a_n+1=sqrt3a_n<3$
(I added the $0<$ just to make absolutely certain that the square root is allowed.)
answered Mar 28 at 20:00
ArthurArthur
121k7122208
$endgroup$
Here is the induction step spelled out.
Assuming $0<a_n<3$, show that $0<a_n+1=sqrt3a_n<3$
(I added the $0<$ just to make absolutely certain that the square root is allowed.)
answered Mar 28 at 20:00
ArthurArthur
121k7122208
answered Mar 28 at 20:00
ArthurArthur
121k7122208
answered Mar 28 at 20:00
ArthurArthur
121k7122208
answered Mar 28 at 20:00
ArthurArthur
121k7122208
121k7122208
add a comment |
add a comment |
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