MHB Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7

  • Thread starter Thread starter simcan18
  • Start date Start date
  • Tags Tags
    Induction
simcan18
Messages
6
Reaction score
0
Can someone with understanding of proof by induction help with this problem?

Prove by induction that 3 raised to 2n+1 + 2 raised to n-1 is divisible by 7 for all numbers greater than/or equal to 1. How do you do the inductive step?
 
Mathematics news on Phys.org
I have done the base case and some of the inductive..which I'm not sure I'm going in the right direction.
Inductive, So does it hold true for n=k+1
3 raised 2(k+1)+1 +2 raised(k+1)-1 = 3 raised 2k+2+1 +2 raised (k+1)-1
= 3 raised 2k+1 x 3 raised2 + 2 raised k x 2 raised 0
=9 x 3 raised 2k+1 + 1 x 2 raised k
= 27 x 3 raised 2k +1x2 raised k

Problem isn't posting correctly
 
$3^{2(k+1)+1}+2^{(k+1)-1}$​

$=\quad3^{2k+3}+2^k$

$=\quad9\cdot3^{2k+1}+2\cdot2^{k-1}$

$=\quad7\cdot3^{2k+1}+2\cdot\left(3^{2k+1}+2^{k-1}\right).$

It should be straightforward to proceed from here.

When doing problems of this kind, look at the number you want your expression to be divisible by (in this case $7$) and try and rearrange your expression to involve it.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top