Proof by induction, inductive step problem
Prove by induction:
$$sum_{i=0}^n 3^i =frac {1}{2} (3^{n+1}-1)$$
Now, i know how to do the first step and i understand it but then i have a problem with the second step which is showing that its true for n+1.
My question is:
Is this notation corect:
$$sum_{i=0}^{n+1} 3^i =frac {1}{2} (3^{n+2}-1)$$
Which of these is corect and why?
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^i+n+1$$
or
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^{n+1}$$
proof-explanation
asked Jan 4 at 20:41
mBartmBart
82
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Prove by induction:
$$sum_{i=0}^n 3^i =frac {1}{2} (3^{n+1}-1)$$
Now, i know how to do the first step and i understand it but then i have a problem with the second step which is showing that its true for n+1.
My question is:
Is this notation corect:
$$sum_{i=0}^{n+1} 3^i =frac {1}{2} (3^{n+2}-1)$$
Which of these is corect and why?
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^i+n+1$$
or
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^{n+1}$$
proof-explanation
asked Jan 4 at 20:41
mBartmBart
82
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Prove by induction:
$$sum_{i=0}^n 3^i =frac {1}{2} (3^{n+1}-1)$$
Now, i know how to do the first step and i understand it but then i have a problem with the second step which is showing that its true for n+1.
My question is:
Is this notation corect:
$$sum_{i=0}^{n+1} 3^i =frac {1}{2} (3^{n+2}-1)$$
Which of these is corect and why?
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^i+n+1$$
or
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^{n+1}$$
proof-explanation
asked Jan 4 at 20:41
mBartmBart
82
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Prove by induction:
$$sum_{i=0}^n 3^i =frac {1}{2} (3^{n+1}-1)$$
Now, i know how to do the first step and i understand it but then i have a problem with the second step which is showing that its true for n+1.
My question is:
Is this notation corect:
$$sum_{i=0}^{n+1} 3^i =frac {1}{2} (3^{n+2}-1)$$
Which of these is corect and why?
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^i+n+1$$
or
$$sum_{i=0}^{n+1} 3^i =sum_{i=0}^{n} 3^{n+1}$$
proof-explanation
proof-explanation
asked Jan 4 at 20:41
mBartmBart
82
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 20:41
mBartmBart
82
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 20:41
mBartmBart
82
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 20:41
mBartmBart
82
asked Jan 4 at 20:41
mBartmBart
82
82
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mBart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
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Neither of those is correct, but the first one is closer. What you want to say is
$$sum_{i=0}^{n+1}3^{i}=sum_{i=0}^{n}3^{i}+3^{n+1}$$
since $3^{n+1}$ is the extra term missing from the sum on the right hand side. To complete the proof, the induction hypothesis implies that the right hand side is
$$frac{1}{2}(3^{n+1}-1)+3^{n+1}=frac{3}{2}cdot3^{n+1}-frac{1}{2}=frac{1}{2}cdot 3^{n+2}-frac{1}{2}=frac{1}{2}(3^{n+2}-1)$$
answered Jan 4 at 20:50
pwerthpwerth
1,977413
add a comment |
Notation looks good.
Your base case is:
$$sum_{i=0}^0{3^i}=3^0=1$$
$$frac12(3^1-1)=frac12(2)=1$$
so true for $n=0$
Inductive Step:
Assume the statement true for $n=k$, that is:
$$sum_{i=0}^k{3^i}=frac12(3^{k+1}-1)$$
Then show for $n=k+1$. We use that:
$$sum_{i=0}^{k+1}{3^i}=sum_{i=0}^{k}{3^i}+3^{k+1}$$
This is the correct expansion of that sum as opposed to what you suggested. We achieve this by finding the $(k+1)$th term of the sum and adding it to the summation to $k$ terms.
So you need to show that:
$$frac12(3^{k+1}-1)+3^{k+1}=frac12(3^{k+2}-1)$$
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
Neither of those is correct, but the first one is closer. What you want to say is
$$sum_{i=0}^{n+1}3^{i}=sum_{i=0}^{n}3^{i}+3^{n+1}$$
since $3^{n+1}$ is the extra term missing from the sum on the right hand side. To complete the proof, the induction hypothesis implies that the right hand side is
$$frac{1}{2}(3^{n+1}-1)+3^{n+1}=frac{3}{2}cdot3^{n+1}-frac{1}{2}=frac{1}{2}cdot 3^{n+2}-frac{1}{2}=frac{1}{2}(3^{n+2}-1)$$
answered Jan 4 at 20:50
pwerthpwerth
1,977413
add a comment |
Neither of those is correct, but the first one is closer. What you want to say is
$$sum_{i=0}^{n+1}3^{i}=sum_{i=0}^{n}3^{i}+3^{n+1}$$
since $3^{n+1}$ is the extra term missing from the sum on the right hand side. To complete the proof, the induction hypothesis implies that the right hand side is
$$frac{1}{2}(3^{n+1}-1)+3^{n+1}=frac{3}{2}cdot3^{n+1}-frac{1}{2}=frac{1}{2}cdot 3^{n+2}-frac{1}{2}=frac{1}{2}(3^{n+2}-1)$$
answered Jan 4 at 20:50
pwerthpwerth
1,977413
add a comment |
Neither of those is correct, but the first one is closer. What you want to say is
$$sum_{i=0}^{n+1}3^{i}=sum_{i=0}^{n}3^{i}+3^{n+1}$$
since $3^{n+1}$ is the extra term missing from the sum on the right hand side. To complete the proof, the induction hypothesis implies that the right hand side is
$$frac{1}{2}(3^{n+1}-1)+3^{n+1}=frac{3}{2}cdot3^{n+1}-frac{1}{2}=frac{1}{2}cdot 3^{n+2}-frac{1}{2}=frac{1}{2}(3^{n+2}-1)$$
answered Jan 4 at 20:50
pwerthpwerth
1,977413
Neither of those is correct, but the first one is closer. What you want to say is
$$sum_{i=0}^{n+1}3^{i}=sum_{i=0}^{n}3^{i}+3^{n+1}$$
since $3^{n+1}$ is the extra term missing from the sum on the right hand side. To complete the proof, the induction hypothesis implies that the right hand side is
$$frac{1}{2}(3^{n+1}-1)+3^{n+1}=frac{3}{2}cdot3^{n+1}-frac{1}{2}=frac{1}{2}cdot 3^{n+2}-frac{1}{2}=frac{1}{2}(3^{n+2}-1)$$
answered Jan 4 at 20:50
pwerthpwerth
1,977413
answered Jan 4 at 20:50
pwerthpwerth
1,977413
answered Jan 4 at 20:50
pwerthpwerth
1,977413
answered Jan 4 at 20:50
pwerthpwerth
1,977413
1,977413
add a comment |
add a comment |
Notation looks good.
Your base case is:
$$sum_{i=0}^0{3^i}=3^0=1$$
$$frac12(3^1-1)=frac12(2)=1$$
so true for $n=0$
Inductive Step:
Assume the statement true for $n=k$, that is:
$$sum_{i=0}^k{3^i}=frac12(3^{k+1}-1)$$
Then show for $n=k+1$. We use that:
$$sum_{i=0}^{k+1}{3^i}=sum_{i=0}^{k}{3^i}+3^{k+1}$$
This is the correct expansion of that sum as opposed to what you suggested. We achieve this by finding the $(k+1)$th term of the sum and adding it to the summation to $k$ terms.
So you need to show that:
$$frac12(3^{k+1}-1)+3^{k+1}=frac12(3^{k+2}-1)$$
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
add a comment |
Notation looks good.
Your base case is:
$$sum_{i=0}^0{3^i}=3^0=1$$
$$frac12(3^1-1)=frac12(2)=1$$
so true for $n=0$
Inductive Step:
Assume the statement true for $n=k$, that is:
$$sum_{i=0}^k{3^i}=frac12(3^{k+1}-1)$$
Then show for $n=k+1$. We use that:
$$sum_{i=0}^{k+1}{3^i}=sum_{i=0}^{k}{3^i}+3^{k+1}$$
This is the correct expansion of that sum as opposed to what you suggested. We achieve this by finding the $(k+1)$th term of the sum and adding it to the summation to $k$ terms.
So you need to show that:
$$frac12(3^{k+1}-1)+3^{k+1}=frac12(3^{k+2}-1)$$
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
add a comment |
Notation looks good.
Your base case is:
$$sum_{i=0}^0{3^i}=3^0=1$$
$$frac12(3^1-1)=frac12(2)=1$$
so true for $n=0$
Inductive Step:
Assume the statement true for $n=k$, that is:
$$sum_{i=0}^k{3^i}=frac12(3^{k+1}-1)$$
Then show for $n=k+1$. We use that:
$$sum_{i=0}^{k+1}{3^i}=sum_{i=0}^{k}{3^i}+3^{k+1}$$
This is the correct expansion of that sum as opposed to what you suggested. We achieve this by finding the $(k+1)$th term of the sum and adding it to the summation to $k$ terms.
So you need to show that:
$$frac12(3^{k+1}-1)+3^{k+1}=frac12(3^{k+2}-1)$$
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
Notation looks good.
Your base case is:
$$sum_{i=0}^0{3^i}=3^0=1$$
$$frac12(3^1-1)=frac12(2)=1$$
so true for $n=0$
Inductive Step:
Assume the statement true for $n=k$, that is:
$$sum_{i=0}^k{3^i}=frac12(3^{k+1}-1)$$
Then show for $n=k+1$. We use that:
$$sum_{i=0}^{k+1}{3^i}=sum_{i=0}^{k}{3^i}+3^{k+1}$$
This is the correct expansion of that sum as opposed to what you suggested. We achieve this by finding the $(k+1)$th term of the sum and adding it to the summation to $k$ terms.
So you need to show that:
$$frac12(3^{k+1}-1)+3^{k+1}=frac12(3^{k+2}-1)$$
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
answered Jan 4 at 20:52
Rhys HughesRhys Hughes
5,1071427
5,1071427
add a comment |
add a comment |
mBart is a new contributor. Be nice, and check out our Code of Conduct.
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