Why is this equality wrong? Can't find a proof
$begingroup$
I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$
Edit: They are equal, thanks for the help
summation geometric-progressions
edited 26 mins ago
Arnaud D.
15.8k52443
asked 2 hours ago
Jack MöllerJack Möller
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$
Edit: They are equal, thanks for the help
summation geometric-progressions
edited 26 mins ago
Arnaud D.
15.8k52443
asked 2 hours ago
Jack MöllerJack Möller
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
2 hours ago
$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
2 hours ago
$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
2 hours ago
$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago
$begingroup$
See math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn
$endgroup$
– Arnaud D.
18 mins ago
add a comment |
$begingroup$
I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$
Edit: They are equal, thanks for the help
summation geometric-progressions
edited 26 mins ago
Arnaud D.
15.8k52443
asked 2 hours ago
Jack MöllerJack Möller
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I can't understant why these two formulas are not equal, I am certain they are but I cannot find a proof for my claim:
$$sum_{i=0}^{x-1} 2^{i} = 2^x-1$$
Edit: They are equal, thanks for the help
summation geometric-progressions
summation geometric-progressions
edited 26 mins ago
Arnaud D.
15.8k52443
asked 2 hours ago
Jack MöllerJack Möller
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 26 mins ago
Arnaud D.
15.8k52443
asked 2 hours ago
Jack MöllerJack Möller
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 26 mins ago
Arnaud D.
15.8k52443
edited 26 mins ago
Arnaud D.
15.8k52443
edited 26 mins ago
Arnaud D.
15.8k52443
15.8k52443
asked 2 hours ago
Jack MöllerJack Möller
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Jack MöllerJack Möller
42
asked 2 hours ago
Jack MöllerJack Möller
42
42
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jack Möller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
2 hours ago
$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
2 hours ago
$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
2 hours ago
$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago
$begingroup$
See math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn
$endgroup$
– Arnaud D.
18 mins ago
add a comment |
3
$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
2 hours ago
$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
2 hours ago
$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
2 hours ago
$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago
$begingroup$
See math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn
$endgroup$
– Arnaud D.
18 mins ago
3
3
$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
2 hours ago
$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
2 hours ago
$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
2 hours ago
$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
2 hours ago
$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
2 hours ago
$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
2 hours ago
$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago
$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago
$begingroup$
See math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn
$endgroup$
– Arnaud D.
18 mins ago
$begingroup$
See math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn
$endgroup$
– Arnaud D.
18 mins ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The easiest way is to notice that
$$
sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
$$
which is a telescoping series.
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
$endgroup$
add a comment |
$begingroup$
$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.
answered 2 hours ago
AndreasAndreas
7,8581037
$endgroup$
add a comment |
$begingroup$
Applying the finite geometric series formula we obtain
begin{align*}
sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
end{align*}
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
$endgroup$
add a comment |
$begingroup$
Use induction to prove it:
Show it is true for $x=1$
Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$
Then show that it is true for $n=k+1$ by using the above formula.
answered 2 hours ago
Ben CrossleyBen Crossley
787318
$endgroup$
add a comment |
$begingroup$
You can prove this equality with the following method:
x belongs to N*
For x=1, 2^0 = 2^1 - 1 = 1
For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
.
.
.
Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),
Proof
Thus, the equality holds for all x in N*
answered 1 hour ago
Wane MamadouWane Mamadou
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The easiest way is to notice that
$$
sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
$$
which is a telescoping series.
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
$endgroup$
add a comment |
$begingroup$
The easiest way is to notice that
$$
sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
$$
which is a telescoping series.
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
$endgroup$
add a comment |
$begingroup$
The easiest way is to notice that
$$
sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
$$
which is a telescoping series.
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
$endgroup$
The easiest way is to notice that
$$
sum_{i=0}^{x-1}2^i=sum_{i=0}^{x-1}(2^{i+1}-2^{i})=({2^1}-2^0)+(2^2-2^1)+(2^3-2^2)+dotsb+(2^x-2^{x-1})
$$
which is a telescoping series.
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
answered 2 hours ago
Foobaz JohnFoobaz John
21.7k41352
21.7k41352
add a comment |
add a comment |
$begingroup$
$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.
answered 2 hours ago
AndreasAndreas
7,8581037
$endgroup$
add a comment |
$begingroup$
$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.
answered 2 hours ago
AndreasAndreas
7,8581037
$endgroup$
add a comment |
$begingroup$
$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.
answered 2 hours ago
AndreasAndreas
7,8581037
$endgroup$
$$
S(x) = sum_{i=0}^{x-1} 2^{i} \
2 S(x) = 2 sum_{i=0}^{x-1} 2^{i} = sum_{i=0}^{x-1} 2^{i+1} = sum_{k=1}^{x} 2^{k} = S(x) + 2^x - 1
$$
so this is your proof that indeed $ S(x) = 2^x - 1 $.
answered 2 hours ago
AndreasAndreas
7,8581037
edited 1 hour ago
answered 2 hours ago
AndreasAndreas
7,8581037
answered 2 hours ago
AndreasAndreas
7,8581037
answered 2 hours ago
AndreasAndreas
7,8581037
7,8581037
add a comment |
add a comment |
$begingroup$
Applying the finite geometric series formula we obtain
begin{align*}
sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
end{align*}
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
$endgroup$
add a comment |
$begingroup$
Applying the finite geometric series formula we obtain
begin{align*}
sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
end{align*}
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
$endgroup$
add a comment |
$begingroup$
Applying the finite geometric series formula we obtain
begin{align*}
sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
end{align*}
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
$endgroup$
Applying the finite geometric series formula we obtain
begin{align*}
sum_{i=0}^{x-1}2^i=frac{2^x-1}{2-1}=2^x-1
end{align*}
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
answered 43 mins ago
Markus ScheuerMarkus Scheuer
60.6k455145
60.6k455145
add a comment |
add a comment |
$begingroup$
Use induction to prove it:
Show it is true for $x=1$
Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$
Then show that it is true for $n=k+1$ by using the above formula.
answered 2 hours ago
Ben CrossleyBen Crossley
787318
$endgroup$
add a comment |
$begingroup$
Use induction to prove it:
Show it is true for $x=1$
Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$
Then show that it is true for $n=k+1$ by using the above formula.
answered 2 hours ago
Ben CrossleyBen Crossley
787318
$endgroup$
add a comment |
$begingroup$
Use induction to prove it:
Show it is true for $x=1$
Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$
Then show that it is true for $n=k+1$ by using the above formula.
answered 2 hours ago
Ben CrossleyBen Crossley
787318
$endgroup$
Use induction to prove it:
Show it is true for $x=1$
Assume it is true for $x=k$, this means that you are allowed to use $$sum_{i=0}^{k-1} 2^{i} = 2^k-1$$
Then show that it is true for $n=k+1$ by using the above formula.
answered 2 hours ago
Ben CrossleyBen Crossley
787318
answered 2 hours ago
Ben CrossleyBen Crossley
787318
answered 2 hours ago
Ben CrossleyBen Crossley
787318
answered 2 hours ago
Ben CrossleyBen Crossley
787318
787318
add a comment |
add a comment |
$begingroup$
You can prove this equality with the following method:
x belongs to N*
For x=1, 2^0 = 2^1 - 1 = 1
For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
.
.
.
Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),
Proof
Thus, the equality holds for all x in N*
answered 1 hour ago
Wane MamadouWane Mamadou
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You can prove this equality with the following method:
x belongs to N*
For x=1, 2^0 = 2^1 - 1 = 1
For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
.
.
.
Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),
Proof
Thus, the equality holds for all x in N*
answered 1 hour ago
Wane MamadouWane Mamadou
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You can prove this equality with the following method:
x belongs to N*
For x=1, 2^0 = 2^1 - 1 = 1
For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
.
.
.
Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),
Proof
Thus, the equality holds for all x in N*
answered 1 hour ago
Wane MamadouWane Mamadou
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You can prove this equality with the following method:
x belongs to N*
For x=1, 2^0 = 2^1 - 1 = 1
For x=2, 2^0 + 2^1 = 2^2 - 1 = 3
.
.
.
Assume that the equality is correct up to (x-1) and demonstrate for the next order (see attached image),
Proof
Thus, the equality holds for all x in N*
answered 1 hour ago
Wane MamadouWane Mamadou
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Wane MamadouWane Mamadou
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Wane MamadouWane Mamadou
1
answered 1 hour ago
Wane MamadouWane Mamadou
1
1
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Wane Mamadou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wane Mamadou 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 |
Jack Möller is a new contributor. Be nice, and check out our Code of Conduct.
Jack Möller is a new contributor. Be nice, and check out our Code of Conduct.
Jack Möller is a new contributor. Be nice, and check out our Code of Conduct.
Jack Möller is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Prove by induction on $xgeq 1$.
$endgroup$
– Wuestenfux
2 hours ago
$begingroup$
Did you really mean to say "(...) are not equal, I am certain they are"?
$endgroup$
– StackTD
2 hours ago
$begingroup$
Multiply the left hand side by $(2-1)$ using the distributive law, simplify.
$endgroup$
– Ned
2 hours ago
$begingroup$
If you are familiar with geometric progressions, you could also use $sum_limits{i = 0}^{x-1} u_1r^i = frac{u_1left(1-r^xright)}{1-r}$.
$endgroup$
– KM101
1 hour ago
$begingroup$
See math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn
$endgroup$
– Arnaud D.
18 mins ago