Prove $sum_{i=1}^n a_i$ = $sum_{i=2}^{n+1} a_{i-1}$
Given $sum_{i=1}^n a_i$ = $sum_{i=2}^{n+1} a_{i-1}$
How would you show this true for all n ∈ N and $a_1, a_2, . . . , a_n$ ∈ R?
I know it is obviously true because i would just use a substitution like i=j-1 then summing j-1 from 2 to n+1 gives the same result but not really sure how to show this. It seems like an induction problem to me but im not sure, how is this done? Thanks
summation
edited Dec 3 '18 at 9:10
Arjang
5,56862363
asked Dec 3 '18 at 9:02
Harry
253
add a comment |
Given $sum_{i=1}^n a_i$ = $sum_{i=2}^{n+1} a_{i-1}$
How would you show this true for all n ∈ N and $a_1, a_2, . . . , a_n$ ∈ R?
I know it is obviously true because i would just use a substitution like i=j-1 then summing j-1 from 2 to n+1 gives the same result but not really sure how to show this. It seems like an induction problem to me but im not sure, how is this done? Thanks
summation
edited Dec 3 '18 at 9:10
Arjang
5,56862363
asked Dec 3 '18 at 9:02
Harry
253
1
The substitution works.
– xbh
Dec 3 '18 at 9:09
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 3 '18 at 9:09
add a comment |
Given $sum_{i=1}^n a_i$ = $sum_{i=2}^{n+1} a_{i-1}$
How would you show this true for all n ∈ N and $a_1, a_2, . . . , a_n$ ∈ R?
I know it is obviously true because i would just use a substitution like i=j-1 then summing j-1 from 2 to n+1 gives the same result but not really sure how to show this. It seems like an induction problem to me but im not sure, how is this done? Thanks
summation
edited Dec 3 '18 at 9:10
Arjang
5,56862363
asked Dec 3 '18 at 9:02
Harry
253
Given $sum_{i=1}^n a_i$ = $sum_{i=2}^{n+1} a_{i-1}$
How would you show this true for all n ∈ N and $a_1, a_2, . . . , a_n$ ∈ R?
I know it is obviously true because i would just use a substitution like i=j-1 then summing j-1 from 2 to n+1 gives the same result but not really sure how to show this. It seems like an induction problem to me but im not sure, how is this done? Thanks
summation
summation
edited Dec 3 '18 at 9:10
Arjang
5,56862363
asked Dec 3 '18 at 9:02
Harry
253
edited Dec 3 '18 at 9:10
Arjang
5,56862363
asked Dec 3 '18 at 9:02
Harry
253
edited Dec 3 '18 at 9:10
Arjang
5,56862363
edited Dec 3 '18 at 9:10
Arjang
5,56862363
edited Dec 3 '18 at 9:10
Arjang
5,56862363
5,56862363
asked Dec 3 '18 at 9:02
Harry
253
asked Dec 3 '18 at 9:02
Harry
253
asked Dec 3 '18 at 9:02
Harry
253
253
1
The substitution works.
– xbh
Dec 3 '18 at 9:09
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 3 '18 at 9:09
add a comment |
1
The substitution works.
– xbh
Dec 3 '18 at 9:09
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 3 '18 at 9:09
1
1
The substitution works.
– xbh
Dec 3 '18 at 9:09
The substitution works.
– xbh
Dec 3 '18 at 9:09
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 3 '18 at 9:09
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 3 '18 at 9:09
add a comment |
2 Answers
2
active
oldest
votes
Yes, you can prove the claim by induction. For that, you must first analyse what the claim even is. In your case, you only have one variable, $n$ ($i$ is bound inside the sum), so clearly, induction will run on $n$, and the claim $P(n)$ will be
$$sum_{i=1}^n a_i = sum_{i=2}^{n+1} a_{i-1}$$
Also, for this proof, you will need to remember the definition of what $$sum_{i=1}^n a_i$$ is. Remember that this expression is defined recursively, i.e.
- $sum_{i=1}^1 a_i = a_1$
- $sum_{i=1}^n a_i = sum_{i=1}^{n-1} a_i + a_n$
The induction has two parts.
Part 1:
You need to prove $P(1)$. This part should be really easy. All you have to do is correctly write out what $P(1)$ means, and use the definitions of sums. In particular, you will use property 1 from above in this step.
Part 2:
Here, you assume that $P(n)$ is true, and you prove the statement $P(n+1)$. Again, correctly writing out $P(n)$ and $P(n+1)$ will get you 99% of the way there. In particular, in this step, property 2 from above will be useful.
answered Dec 3 '18 at 9:11
5xum
89.6k393161
add a comment |
Prove that
$F(n):= sum_{i=1}^{n}a_i-sum_{i=2}^{n+1}a_{i-1}= 0$ , for $n in mathbb{Z^+}$ by induction.
1) $n=1$√.
2) Hypothesis $F(n)=0$.
3) Step for $n+1$.
$F(n+1)=$
$sum_{i=1}^{n+1}a_i - sum_{i=2}^{n+2}a_{i-1}=$
$sum_{i=1}^{n}a_i +a_{n+1}$
$- sum_{i=2}^{n+1}a_{i-1}- a_{n+1}=$
$F(n)+(a_{n+1}-a_{n+1})=0$,
since $F(n)=0$ by hypothesis, and the second summand is zero.
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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votes
Yes, you can prove the claim by induction. For that, you must first analyse what the claim even is. In your case, you only have one variable, $n$ ($i$ is bound inside the sum), so clearly, induction will run on $n$, and the claim $P(n)$ will be
$$sum_{i=1}^n a_i = sum_{i=2}^{n+1} a_{i-1}$$
Also, for this proof, you will need to remember the definition of what $$sum_{i=1}^n a_i$$ is. Remember that this expression is defined recursively, i.e.
- $sum_{i=1}^1 a_i = a_1$
- $sum_{i=1}^n a_i = sum_{i=1}^{n-1} a_i + a_n$
The induction has two parts.
Part 1:
You need to prove $P(1)$. This part should be really easy. All you have to do is correctly write out what $P(1)$ means, and use the definitions of sums. In particular, you will use property 1 from above in this step.
Part 2:
Here, you assume that $P(n)$ is true, and you prove the statement $P(n+1)$. Again, correctly writing out $P(n)$ and $P(n+1)$ will get you 99% of the way there. In particular, in this step, property 2 from above will be useful.
answered Dec 3 '18 at 9:11
5xum
89.6k393161
add a comment |
Yes, you can prove the claim by induction. For that, you must first analyse what the claim even is. In your case, you only have one variable, $n$ ($i$ is bound inside the sum), so clearly, induction will run on $n$, and the claim $P(n)$ will be
$$sum_{i=1}^n a_i = sum_{i=2}^{n+1} a_{i-1}$$
Also, for this proof, you will need to remember the definition of what $$sum_{i=1}^n a_i$$ is. Remember that this expression is defined recursively, i.e.
- $sum_{i=1}^1 a_i = a_1$
- $sum_{i=1}^n a_i = sum_{i=1}^{n-1} a_i + a_n$
The induction has two parts.
Part 1:
You need to prove $P(1)$. This part should be really easy. All you have to do is correctly write out what $P(1)$ means, and use the definitions of sums. In particular, you will use property 1 from above in this step.
Part 2:
Here, you assume that $P(n)$ is true, and you prove the statement $P(n+1)$. Again, correctly writing out $P(n)$ and $P(n+1)$ will get you 99% of the way there. In particular, in this step, property 2 from above will be useful.
answered Dec 3 '18 at 9:11
5xum
89.6k393161
add a comment |
Yes, you can prove the claim by induction. For that, you must first analyse what the claim even is. In your case, you only have one variable, $n$ ($i$ is bound inside the sum), so clearly, induction will run on $n$, and the claim $P(n)$ will be
$$sum_{i=1}^n a_i = sum_{i=2}^{n+1} a_{i-1}$$
Also, for this proof, you will need to remember the definition of what $$sum_{i=1}^n a_i$$ is. Remember that this expression is defined recursively, i.e.
- $sum_{i=1}^1 a_i = a_1$
- $sum_{i=1}^n a_i = sum_{i=1}^{n-1} a_i + a_n$
The induction has two parts.
Part 1:
You need to prove $P(1)$. This part should be really easy. All you have to do is correctly write out what $P(1)$ means, and use the definitions of sums. In particular, you will use property 1 from above in this step.
Part 2:
Here, you assume that $P(n)$ is true, and you prove the statement $P(n+1)$. Again, correctly writing out $P(n)$ and $P(n+1)$ will get you 99% of the way there. In particular, in this step, property 2 from above will be useful.
answered Dec 3 '18 at 9:11
5xum
89.6k393161
Yes, you can prove the claim by induction. For that, you must first analyse what the claim even is. In your case, you only have one variable, $n$ ($i$ is bound inside the sum), so clearly, induction will run on $n$, and the claim $P(n)$ will be
$$sum_{i=1}^n a_i = sum_{i=2}^{n+1} a_{i-1}$$
Also, for this proof, you will need to remember the definition of what $$sum_{i=1}^n a_i$$ is. Remember that this expression is defined recursively, i.e.
- $sum_{i=1}^1 a_i = a_1$
- $sum_{i=1}^n a_i = sum_{i=1}^{n-1} a_i + a_n$
The induction has two parts.
Part 1:
You need to prove $P(1)$. This part should be really easy. All you have to do is correctly write out what $P(1)$ means, and use the definitions of sums. In particular, you will use property 1 from above in this step.
Part 2:
Here, you assume that $P(n)$ is true, and you prove the statement $P(n+1)$. Again, correctly writing out $P(n)$ and $P(n+1)$ will get you 99% of the way there. In particular, in this step, property 2 from above will be useful.
answered Dec 3 '18 at 9:11
5xum
89.6k393161
answered Dec 3 '18 at 9:11
5xum
89.6k393161
answered Dec 3 '18 at 9:11
5xum
89.6k393161
answered Dec 3 '18 at 9:11
5xum
89.6k393161
89.6k393161
add a comment |
add a comment |
Prove that
$F(n):= sum_{i=1}^{n}a_i-sum_{i=2}^{n+1}a_{i-1}= 0$ , for $n in mathbb{Z^+}$ by induction.
1) $n=1$√.
2) Hypothesis $F(n)=0$.
3) Step for $n+1$.
$F(n+1)=$
$sum_{i=1}^{n+1}a_i - sum_{i=2}^{n+2}a_{i-1}=$
$sum_{i=1}^{n}a_i +a_{n+1}$
$- sum_{i=2}^{n+1}a_{i-1}- a_{n+1}=$
$F(n)+(a_{n+1}-a_{n+1})=0$,
since $F(n)=0$ by hypothesis, and the second summand is zero.
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
add a comment |
Prove that
$F(n):= sum_{i=1}^{n}a_i-sum_{i=2}^{n+1}a_{i-1}= 0$ , for $n in mathbb{Z^+}$ by induction.
1) $n=1$√.
2) Hypothesis $F(n)=0$.
3) Step for $n+1$.
$F(n+1)=$
$sum_{i=1}^{n+1}a_i - sum_{i=2}^{n+2}a_{i-1}=$
$sum_{i=1}^{n}a_i +a_{n+1}$
$- sum_{i=2}^{n+1}a_{i-1}- a_{n+1}=$
$F(n)+(a_{n+1}-a_{n+1})=0$,
since $F(n)=0$ by hypothesis, and the second summand is zero.
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
add a comment |
Prove that
$F(n):= sum_{i=1}^{n}a_i-sum_{i=2}^{n+1}a_{i-1}= 0$ , for $n in mathbb{Z^+}$ by induction.
1) $n=1$√.
2) Hypothesis $F(n)=0$.
3) Step for $n+1$.
$F(n+1)=$
$sum_{i=1}^{n+1}a_i - sum_{i=2}^{n+2}a_{i-1}=$
$sum_{i=1}^{n}a_i +a_{n+1}$
$- sum_{i=2}^{n+1}a_{i-1}- a_{n+1}=$
$F(n)+(a_{n+1}-a_{n+1})=0$,
since $F(n)=0$ by hypothesis, and the second summand is zero.
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
Prove that
$F(n):= sum_{i=1}^{n}a_i-sum_{i=2}^{n+1}a_{i-1}= 0$ , for $n in mathbb{Z^+}$ by induction.
1) $n=1$√.
2) Hypothesis $F(n)=0$.
3) Step for $n+1$.
$F(n+1)=$
$sum_{i=1}^{n+1}a_i - sum_{i=2}^{n+2}a_{i-1}=$
$sum_{i=1}^{n}a_i +a_{n+1}$
$- sum_{i=2}^{n+1}a_{i-1}- a_{n+1}=$
$F(n)+(a_{n+1}-a_{n+1})=0$,
since $F(n)=0$ by hypothesis, and the second summand is zero.
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
answered Dec 3 '18 at 9:44
Peter Szilas
10.7k2720
10.7k2720
add a comment |
add a comment |
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1
The substitution works.
– xbh
Dec 3 '18 at 9:09
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Dec 3 '18 at 9:09