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?












0












$begingroup$


I discovered this yesterday and I am wondering:




  1. Is this a well known formula?

  2. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


I discovered this yesterday and I am wondering:




  1. Is this a well known formula?

  2. 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


I discovered this yesterday and I am wondering:




  1. Is this a well known formula?

  2. 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







share|cite|improve this question











$endgroup$




I discovered this yesterday and I am wondering:




  1. Is this a well known formula?

  2. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 7:11







ReverseFlow

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















3












$begingroup$

It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












$begingroup$

It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$






share|cite|improve this answer









$endgroup$



It is:
$$3^n+sum_{k=0}^{n-1}3^{n-k-1}2^{2k}=3^n+3^{n-1}sum_{k=0}^{n-1}left(frac43right)^{k}=3^n+3^{n-1}frac{left(frac43right)^n-1}{frac43-1}=\
3^n+(4^n-3^n)=4^n=2^{2n}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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