Mixing
Using Induction to prove this inequality?
$begingroup$
I'm having a hard time solving the following problem:
Let ${displaystyle (a_{n})_{nin mathbb {N_{0}} }}$ be the sequence defined recursively by:
$a_{0}=2$
$a_{n+1} = frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i qquad forall n in mathbb{N_{0}}$
Prove (by induction):
$$a_{n} le 3^{n} + 2^nqquad forall n in mathbb{N_{0}} $$
Every way I tried so far gets me nowhere and I'm starting to get confused. Solutions or any hints will be highly appreciated. Thank you in advance.
sequences-and-series induction natural-numbers
asked Jan 20 at 16:40
NebzatNebzat
244
$endgroup$
add a comment |
$begingroup$
I'm having a hard time solving the following problem:
Let ${displaystyle (a_{n})_{nin mathbb {N_{0}} }}$ be the sequence defined recursively by:
$a_{0}=2$
$a_{n+1} = frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i qquad forall n in mathbb{N_{0}}$
Prove (by induction):
$$a_{n} le 3^{n} + 2^nqquad forall n in mathbb{N_{0}} $$
Every way I tried so far gets me nowhere and I'm starting to get confused. Solutions or any hints will be highly appreciated. Thank you in advance.
sequences-and-series induction natural-numbers
asked Jan 20 at 16:40
NebzatNebzat
244
$endgroup$
add a comment |
$begingroup$
I'm having a hard time solving the following problem:
Let ${displaystyle (a_{n})_{nin mathbb {N_{0}} }}$ be the sequence defined recursively by:
$a_{0}=2$
$a_{n+1} = frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i qquad forall n in mathbb{N_{0}}$
Prove (by induction):
$$a_{n} le 3^{n} + 2^nqquad forall n in mathbb{N_{0}} $$
Every way I tried so far gets me nowhere and I'm starting to get confused. Solutions or any hints will be highly appreciated. Thank you in advance.
sequences-and-series induction natural-numbers
asked Jan 20 at 16:40
NebzatNebzat
244
$endgroup$
I'm having a hard time solving the following problem:
Let ${displaystyle (a_{n})_{nin mathbb {N_{0}} }}$ be the sequence defined recursively by:
$a_{0}=2$
$a_{n+1} = frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i qquad forall n in mathbb{N_{0}}$
Prove (by induction):
$$a_{n} le 3^{n} + 2^nqquad forall n in mathbb{N_{0}} $$
Every way I tried so far gets me nowhere and I'm starting to get confused. Solutions or any hints will be highly appreciated. Thank you in advance.
sequences-and-series induction natural-numbers
sequences-and-series induction natural-numbers
asked Jan 20 at 16:40
NebzatNebzat
244
asked Jan 20 at 16:40
NebzatNebzat
244
edited Jan 20 at 16:48
Nebzat
asked Jan 20 at 16:40
NebzatNebzat
244
asked Jan 20 at 16:40
NebzatNebzat
244
asked Jan 20 at 16:40
NebzatNebzat
244
244
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint:
Using strong induction, you have
begin{align}
a_{n+1}= frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i&le frac{1+3^{n+1}}{2} + sum_{i=0}^n(3^i+ 2^i)\
& = frac{1+3^{n+1}}{2}+frac{3^{n+1}-1}2 + 2^{n+1}-1.
end{align}
Can you end it?
answered Jan 20 at 16:56
BernardBernard
122k740116
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Using strong induction, you have
begin{align}
a_{n+1}= frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i&le frac{1+3^{n+1}}{2} + sum_{i=0}^n(3^i+ 2^i)\
& = frac{1+3^{n+1}}{2}+frac{3^{n+1}-1}2 + 2^{n+1}-1.
end{align}
Can you end it?
answered Jan 20 at 16:56
BernardBernard
122k740116
$endgroup$
add a comment |
$begingroup$
Hint:
Using strong induction, you have
begin{align}
a_{n+1}= frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i&le frac{1+3^{n+1}}{2} + sum_{i=0}^n(3^i+ 2^i)\
& = frac{1+3^{n+1}}{2}+frac{3^{n+1}-1}2 + 2^{n+1}-1.
end{align}
Can you end it?
answered Jan 20 at 16:56
BernardBernard
122k740116
$endgroup$
add a comment |
$begingroup$
Hint:
Using strong induction, you have
begin{align}
a_{n+1}= frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i&le frac{1+3^{n+1}}{2} + sum_{i=0}^n(3^i+ 2^i)\
& = frac{1+3^{n+1}}{2}+frac{3^{n+1}-1}2 + 2^{n+1}-1.
end{align}
Can you end it?
answered Jan 20 at 16:56
BernardBernard
122k740116
$endgroup$
Hint:
Using strong induction, you have
begin{align}
a_{n+1}= frac{1+3^{n+1}}{2} + sum_{i=0}^n a_i&le frac{1+3^{n+1}}{2} + sum_{i=0}^n(3^i+ 2^i)\
& = frac{1+3^{n+1}}{2}+frac{3^{n+1}-1}2 + 2^{n+1}-1.
end{align}
Can you end it?
answered Jan 20 at 16:56
BernardBernard
122k740116
answered Jan 20 at 16:56
BernardBernard
122k740116
answered Jan 20 at 16:56
BernardBernard
122k740116
answered Jan 20 at 16:56
BernardBernard
122k740116
122k740116
add a comment |
add a comment |
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