Mixing
Proving that equality is applicable to any N number
$a_{n}$ is a sequence of numbers defined by:
$a_{0}$ = −1
$a_{1}$ = 3
$a_{n+2}$ = 6$a_{n}$ − $a_{n+1}$ + 4n + 1
I have to prove that for every natural number n this equality is applicable:
$a_{n}$ = $2^{n}$ − $(-3)^{n}$ − n − 1
Any ideas/hints how to proceed? I have no idea what to do with that.
Sorry for bad English. That is not my first language.
sequences-and-series proof-explanation
asked Nov 21 '18 at 15:10
dfps12
11
add a comment |
$a_{n}$ is a sequence of numbers defined by:
$a_{0}$ = −1
$a_{1}$ = 3
$a_{n+2}$ = 6$a_{n}$ − $a_{n+1}$ + 4n + 1
I have to prove that for every natural number n this equality is applicable:
$a_{n}$ = $2^{n}$ − $(-3)^{n}$ − n − 1
Any ideas/hints how to proceed? I have no idea what to do with that.
Sorry for bad English. That is not my first language.
sequences-and-series proof-explanation
asked Nov 21 '18 at 15:10
dfps12
11
2
If $b_n=2^n-(-3)^n-n-1$ then it would suffice to show that $b_0=a_0, b_1=a_1$ and that $b_n$ also satisfies the recursion $b_{n+2}=6b_n-b_{n+1}+4n+1$ (since that recursion defines a sequence uniquely up to initial conditions).
– lulu
Nov 21 '18 at 15:18
The comment from @lulu is correct, but it might be useful to mention that the parenthetical point at the end of that comment would be proved by strong induction on $n$. Alternatively, without introducing a separate notation like $b_n$, you could directly prove the desired equation by strong induction on $n$.
– Andreas Blass
Nov 21 '18 at 15:54
add a comment |
$a_{n}$ is a sequence of numbers defined by:
$a_{0}$ = −1
$a_{1}$ = 3
$a_{n+2}$ = 6$a_{n}$ − $a_{n+1}$ + 4n + 1
I have to prove that for every natural number n this equality is applicable:
$a_{n}$ = $2^{n}$ − $(-3)^{n}$ − n − 1
Any ideas/hints how to proceed? I have no idea what to do with that.
Sorry for bad English. That is not my first language.
sequences-and-series proof-explanation
asked Nov 21 '18 at 15:10
dfps12
11
$a_{n}$ is a sequence of numbers defined by:
$a_{0}$ = −1
$a_{1}$ = 3
$a_{n+2}$ = 6$a_{n}$ − $a_{n+1}$ + 4n + 1
I have to prove that for every natural number n this equality is applicable:
$a_{n}$ = $2^{n}$ − $(-3)^{n}$ − n − 1
Any ideas/hints how to proceed? I have no idea what to do with that.
Sorry for bad English. That is not my first language.
sequences-and-series proof-explanation
sequences-and-series proof-explanation
asked Nov 21 '18 at 15:10
dfps12
11
asked Nov 21 '18 at 15:10
dfps12
11
asked Nov 21 '18 at 15:10
dfps12
11
asked Nov 21 '18 at 15:10
dfps12
11
asked Nov 21 '18 at 15:10
dfps12
11
11
2
If $b_n=2^n-(-3)^n-n-1$ then it would suffice to show that $b_0=a_0, b_1=a_1$ and that $b_n$ also satisfies the recursion $b_{n+2}=6b_n-b_{n+1}+4n+1$ (since that recursion defines a sequence uniquely up to initial conditions).
– lulu
Nov 21 '18 at 15:18
The comment from @lulu is correct, but it might be useful to mention that the parenthetical point at the end of that comment would be proved by strong induction on $n$. Alternatively, without introducing a separate notation like $b_n$, you could directly prove the desired equation by strong induction on $n$.
– Andreas Blass
Nov 21 '18 at 15:54
add a comment |
2
If $b_n=2^n-(-3)^n-n-1$ then it would suffice to show that $b_0=a_0, b_1=a_1$ and that $b_n$ also satisfies the recursion $b_{n+2}=6b_n-b_{n+1}+4n+1$ (since that recursion defines a sequence uniquely up to initial conditions).
– lulu
Nov 21 '18 at 15:18
The comment from @lulu is correct, but it might be useful to mention that the parenthetical point at the end of that comment would be proved by strong induction on $n$. Alternatively, without introducing a separate notation like $b_n$, you could directly prove the desired equation by strong induction on $n$.
– Andreas Blass
Nov 21 '18 at 15:54
2
2
If $b_n=2^n-(-3)^n-n-1$ then it would suffice to show that $b_0=a_0, b_1=a_1$ and that $b_n$ also satisfies the recursion $b_{n+2}=6b_n-b_{n+1}+4n+1$ (since that recursion defines a sequence uniquely up to initial conditions).
– lulu
Nov 21 '18 at 15:18
If $b_n=2^n-(-3)^n-n-1$ then it would suffice to show that $b_0=a_0, b_1=a_1$ and that $b_n$ also satisfies the recursion $b_{n+2}=6b_n-b_{n+1}+4n+1$ (since that recursion defines a sequence uniquely up to initial conditions).
– lulu
Nov 21 '18 at 15:18
The comment from @lulu is correct, but it might be useful to mention that the parenthetical point at the end of that comment would be proved by strong induction on $n$. Alternatively, without introducing a separate notation like $b_n$, you could directly prove the desired equation by strong induction on $n$.
– Andreas Blass
Nov 21 '18 at 15:54
The comment from @lulu is correct, but it might be useful to mention that the parenthetical point at the end of that comment would be proved by strong induction on $n$. Alternatively, without introducing a separate notation like $b_n$, you could directly prove the desired equation by strong induction on $n$.
– Andreas Blass
Nov 21 '18 at 15:54
add a comment |
1 Answer
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active
oldest
votes
Let's see three consecutive terms of the series and suppose that n is even:
$a_n=2^n - 3^n-n-1$
$a_{n+1}=2^{n+1} + 3^{n+1}-(n+1)-1$
$a_{n+2}=2^{n+2} -3^{n+2}-(n+2)-1$
Form the sum of $a_{n+1}$ and $a_{n+2}$:
$a_{n+1}+a_{n+2}=(2+4)2^n+(3-9)3^n-2n-5=6overbrace{{(2^n-3^n-n-1)}}^{a_n}+4n+1$
$a_{n+2}=6a_n-a_{n+1}+4n+1$
We can do this for odd n, the result is same.
answered Nov 25 '18 at 7:17
JV.Stalker
58339
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
Let's see three consecutive terms of the series and suppose that n is even:
$a_n=2^n - 3^n-n-1$
$a_{n+1}=2^{n+1} + 3^{n+1}-(n+1)-1$
$a_{n+2}=2^{n+2} -3^{n+2}-(n+2)-1$
Form the sum of $a_{n+1}$ and $a_{n+2}$:
$a_{n+1}+a_{n+2}=(2+4)2^n+(3-9)3^n-2n-5=6overbrace{{(2^n-3^n-n-1)}}^{a_n}+4n+1$
$a_{n+2}=6a_n-a_{n+1}+4n+1$
We can do this for odd n, the result is same.
answered Nov 25 '18 at 7:17
JV.Stalker
58339
add a comment |
Let's see three consecutive terms of the series and suppose that n is even:
$a_n=2^n - 3^n-n-1$
$a_{n+1}=2^{n+1} + 3^{n+1}-(n+1)-1$
$a_{n+2}=2^{n+2} -3^{n+2}-(n+2)-1$
Form the sum of $a_{n+1}$ and $a_{n+2}$:
$a_{n+1}+a_{n+2}=(2+4)2^n+(3-9)3^n-2n-5=6overbrace{{(2^n-3^n-n-1)}}^{a_n}+4n+1$
$a_{n+2}=6a_n-a_{n+1}+4n+1$
We can do this for odd n, the result is same.
answered Nov 25 '18 at 7:17
JV.Stalker
58339
add a comment |
Let's see three consecutive terms of the series and suppose that n is even:
$a_n=2^n - 3^n-n-1$
$a_{n+1}=2^{n+1} + 3^{n+1}-(n+1)-1$
$a_{n+2}=2^{n+2} -3^{n+2}-(n+2)-1$
Form the sum of $a_{n+1}$ and $a_{n+2}$:
$a_{n+1}+a_{n+2}=(2+4)2^n+(3-9)3^n-2n-5=6overbrace{{(2^n-3^n-n-1)}}^{a_n}+4n+1$
$a_{n+2}=6a_n-a_{n+1}+4n+1$
We can do this for odd n, the result is same.
answered Nov 25 '18 at 7:17
JV.Stalker
58339
Let's see three consecutive terms of the series and suppose that n is even:
$a_n=2^n - 3^n-n-1$
$a_{n+1}=2^{n+1} + 3^{n+1}-(n+1)-1$
$a_{n+2}=2^{n+2} -3^{n+2}-(n+2)-1$
Form the sum of $a_{n+1}$ and $a_{n+2}$:
$a_{n+1}+a_{n+2}=(2+4)2^n+(3-9)3^n-2n-5=6overbrace{{(2^n-3^n-n-1)}}^{a_n}+4n+1$
$a_{n+2}=6a_n-a_{n+1}+4n+1$
We can do this for odd n, the result is same.
answered Nov 25 '18 at 7:17
JV.Stalker
58339
answered Nov 25 '18 at 7:17
JV.Stalker
58339
answered Nov 25 '18 at 7:17
JV.Stalker
58339
answered Nov 25 '18 at 7:17
JV.Stalker
58339
58339
add a comment |
add a comment |
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If $b_n=2^n-(-3)^n-n-1$ then it would suffice to show that $b_0=a_0, b_1=a_1$ and that $b_n$ also satisfies the recursion $b_{n+2}=6b_n-b_{n+1}+4n+1$ (since that recursion defines a sequence uniquely up to initial conditions).
– lulu
Nov 21 '18 at 15:18
The comment from @lulu is correct, but it might be useful to mention that the parenthetical point at the end of that comment would be proved by strong induction on $n$. Alternatively, without introducing a separate notation like $b_n$, you could directly prove the desired equation by strong induction on $n$.
– Andreas Blass
Nov 21 '18 at 15:54