How to prove a sum formula about powers of 2 and 3? The Next CEO of Stack OverflowProve:...
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How to prove a sum formula about powers of 2 and 3?
The Next CEO of Stack OverflowProve: $sum_{x=0}^{n} (-1)^x {n choose x} = 0$Prove a sum formula by inductionAlternating sum of product of Fibonacci numbersInduction based on sum of $kth$ powers.How can we prove the sum of squares/cubes/etc is always a polynomial of appropriate degree?Formula for sum of first $n$ odd integersProve the theorem by binding the term and splitting the sumHow to prove this summation formula?How can I write this as a double sum?How do I prove this equation by mathematical induction?
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I discovered this yesterday and I am wondering:
- Is this a well known formula?
- How do I go about proving something like this? (Beyond induction.)
$$2^{2n} = 3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}$$
Thank you. Any thoughts on this would be greatly appreciated.
P.s: For notation purposes, when $n=0$ the sum is not evaluated.
combinatorics summation induction exact-sequence
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
$endgroup$
add a comment |
$begingroup$
I discovered this yesterday and I am wondering:
- Is this a well known formula?
- How do I go about proving something like this? (Beyond induction.)
$$2^{2n} = 3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}$$
Thank you. Any thoughts on this would be greatly appreciated.
P.s: For notation purposes, when $n=0$ the sum is not evaluated.
combinatorics summation induction exact-sequence
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
$endgroup$
$begingroup$
Use the geometric sum formula $sum_{k=0}^{n-1} x^k = (1-x^n) / (1-x)$
$endgroup$
– Jair Taylor
Mar 17 at 7:10
$begingroup$
@JairTaylor: it is unclear to me, how that helps.
$endgroup$
– ReverseFlow
Mar 17 at 7:12
$begingroup$
Close, but it does not explain the powers of 2. And after you add in the binomial coeficients, it explains things even less.
$endgroup$
– ReverseFlow
Mar 17 at 7:16
add a comment |
$begingroup$
I discovered this yesterday and I am wondering:
- Is this a well known formula?
- How do I go about proving something like this? (Beyond induction.)
$$2^{2n} = 3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}$$
Thank you. Any thoughts on this would be greatly appreciated.
P.s: For notation purposes, when $n=0$ the sum is not evaluated.
combinatorics summation induction exact-sequence
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
$endgroup$
I discovered this yesterday and I am wondering:
- Is this a well known formula?
- How do I go about proving something like this? (Beyond induction.)
$$2^{2n} = 3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}$$
Thank you. Any thoughts on this would be greatly appreciated.
P.s: For notation purposes, when $n=0$ the sum is not evaluated.
combinatorics summation induction exact-sequence
combinatorics summation induction exact-sequence
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
edited Mar 17 at 7:11
ReverseFlow
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
asked Mar 17 at 7:08
ReverseFlowReverseFlow
609513
609513
$begingroup$
Use the geometric sum formula $sum_{k=0}^{n-1} x^k = (1-x^n) / (1-x)$
$endgroup$
– Jair Taylor
Mar 17 at 7:10
$begingroup$
@JairTaylor: it is unclear to me, how that helps.
$endgroup$
– ReverseFlow
Mar 17 at 7:12
$begingroup$
Close, but it does not explain the powers of 2. And after you add in the binomial coeficients, it explains things even less.
$endgroup$
– ReverseFlow
Mar 17 at 7:16
add a comment |
$begingroup$
Use the geometric sum formula $sum_{k=0}^{n-1} x^k = (1-x^n) / (1-x)$
$endgroup$
– Jair Taylor
Mar 17 at 7:10
$begingroup$
@JairTaylor: it is unclear to me, how that helps.
$endgroup$
– ReverseFlow
Mar 17 at 7:12
$begingroup$
Close, but it does not explain the powers of 2. And after you add in the binomial coeficients, it explains things even less.
$endgroup$
– ReverseFlow
Mar 17 at 7:16
$begingroup$
Use the geometric sum formula $sum_{k=0}^{n-1} x^k = (1-x^n) / (1-x)$
$endgroup$
– Jair Taylor
Mar 17 at 7:10
$begingroup$
Use the geometric sum formula $sum_{k=0}^{n-1} x^k = (1-x^n) / (1-x)$
$endgroup$
– Jair Taylor
Mar 17 at 7:10
$begingroup$
@JairTaylor: it is unclear to me, how that helps.
$endgroup$
– ReverseFlow
Mar 17 at 7:12
$begingroup$
@JairTaylor: it is unclear to me, how that helps.
$endgroup$
– ReverseFlow
Mar 17 at 7:12
$begingroup$
Close, but it does not explain the powers of 2. And after you add in the binomial coeficients, it explains things even less.
$endgroup$
– ReverseFlow
Mar 17 at 7:16
$begingroup$
Close, but it does not explain the powers of 2. And after you add in the binomial coeficients, it explains things even less.
$endgroup$
– ReverseFlow
Mar 17 at 7:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
$endgroup$
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
$endgroup$
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
add a comment |
$begingroup$
It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
$endgroup$
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
add a comment |
$begingroup$
It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
$endgroup$
It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
answered Mar 17 at 7:18
farruhotafarruhota
21.7k2842
21.7k2842
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
add a comment |
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
Beautiful, thank you. Also, I have to wait 4 minutes to accept an answer, -_-.
$endgroup$
– ReverseFlow
Mar 17 at 7:19
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
$begingroup$
No hurry, you should wait longer to see if others can provide more elegant solutions. Good luck.
$endgroup$
– farruhota
Mar 17 at 7:20
add a comment |
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$begingroup$
Use the geometric sum formula $sum_{k=0}^{n-1} x^k = (1-x^n) / (1-x)$
$endgroup$
– Jair Taylor
Mar 17 at 7:10
$begingroup$
@JairTaylor: it is unclear to me, how that helps.
$endgroup$
– ReverseFlow
Mar 17 at 7:12
$begingroup$
Close, but it does not explain the powers of 2. And after you add in the binomial coeficients, it explains things even less.
$endgroup$
– ReverseFlow
Mar 17 at 7:16