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I do not know the rules governing the transformation of each member of the following group of equations into the the next one:

$$ 5(1 + 2^{k -1} + 3^{k -1}) - 6(1 + 2^{k -2} + 3^{k -2}) + 2 \\= (5 -6 + 2) + (5 \times 2^{k - 1} - 6 \times 2^{k - 2}) + (5 \times 3^{k -1} - 6 \times 3^{k - 2}) \\= 1 + (5 - 6 \times 2^{-1}) \times 2^{k -1} + (5 - 6 \times 3^{-1}) \times 3^{k -1} \\= 1 + (5 -3) \times 2^{k-1} + (5 - 2) \times 3^{k -1} \\= 1 + 2 \times 2^{k -1} + 3 \times 3^{k -1} \\= 1 + 2^{k} + 3^{k}$$

I need to learn rules such as these from scratch. Where can I learn them?

Could someone label the rule used to form each member of the group of equalities from the previous member?

Edward.Lin
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  • I know Kahn Academy has some great resources for learning fundamentals. I'm not sure if what I linked is exactly what you want, but if you look around you can definitely find what you're looking for on their website. – HallaSurvivor Jun 09 '21 at 22:11
  • Starting at the top line, you want to combine powers of $2$ and powers of $3$. The general rule is that $a^{r+1} = a \times a^r$. This means that the expression (for example) of $2^{k-1}$ can be re-expressed as $2 \times 2^{k-2}.$ Therefore, the powers of $2$ in the top line can be combined into $$\left(5 \times 2 \times 2^{k-2}\right) - \left(6 \times 2^{k-2}\right) = \left(4 \times 2^{k-2}\right) = 2^k.$$ – user2661923 Jun 09 '21 at 22:11
  • @user2661923 I don't understand your sentence starting "Therefore". It involves an equation manpulation which is not clear to me. – Edward.Lin Jun 09 '21 at 22:16
  • Since $2^{k-1} = \left(2 \times 2^{k-2}\right)$, in the top line, the very first term that involves a power of $2$, namely $\left(5 \times 2^{k-1}\right)$ can be re-expresed as $$\left[5 \times \left(2 \times 2^{k-2}\right) \right].$$ – user2661923 Jun 09 '21 at 22:19
  • The axioms of a field will allow you to derive all of these equations systematically. You can find them here. https://mathworld.wolfram.com/FieldAxioms.html – CyclotomicField Jun 09 '21 at 23:41
  • Check for: Associativity, Commutativity, Distributivity. – Mauro ALLEGRANZA Jun 10 '21 at 06:38
  • @CyclotomicField In https://math.stackexchange.com/questions/675976/proof-by-induction-sequence-of-integers/676012?noredirect=1#comment8633672_676012, the proof has just now been filled in, but solved using principles which are not those of the axioms of a field. So it is unclear to me how the axioms of a field are sufficient to explain the solution given there. – Edward.Lin Jun 10 '21 at 07:36
  • @Edward.Lin a different question would require a different answer. The manipulation of the equations given here only require the field axioms, the question you've linked to requires induction. Change the question and you change the answer. – CyclotomicField Jun 10 '21 at 10:43
  • @CyclotomicField, I can't see how the second line in my question follows from the first, using the field axioms, nor how the third follows from the second. – Edward.Lin Jun 10 '21 at 11:21
  • Worth mentioning that this post relates to your post here. – Will Orrick Jun 10 '21 at 11:29
  • @Edward.Lin the third follows from the second by using distributivity twice, once for $2^{k-1}$ and once for $3^{k-1}$. – CyclotomicField Jun 10 '21 at 14:33

1 Answers1

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Here's an explanation of some of the lines:

  1. This line is obtained from line 1 by bringing the coefficients $5$, $6$ into their respective parentheses using the distributive law and then rearranging and regrouping terms using the commutative and associative laws. I assume there are no issues here, but if so, please ask.
  2. Factors $2^{k-1}$ and $3^{k-1}$ are pulled out of the 2nd and 3rd parentheses, again using the distributive law, but rules of exponents need to be used first. In the 2nd parentheses, the term $-6\cdot2^{k-2}$ has been rewritten as $-6\cdot2^{-1}\cdot2^{k-1}$ using the rule $a^{x+y}=a^xa^y$. Here $a=2$, $x=-1$ and $y=k-1$. In the 3rd parentheses, something similar was done: $-6\cdot3^{k-2}=-6\cdot3^{-1}\cdot3^{k-1}$.
  3. Arithmetic: $6\cdot 2^{-1}=6\cdot\frac{1}{2}=3$. This uses $a^{-x}=\frac{1}{a^x}$ with $a=2$ and $x=1$.
  4. Basic arithmetic.
  5. The law $a^xa^y=a^{x+y}$ is used again here: $2\cdot2^{k-1}=2^1\cdot2^{k-1}=2^{1+k-1}=2^k$ and similarly for the term with base $3$.
Will Orrick
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  • I can't see how line 2 is obtained from 1, using the distributive, commutative and associative laws. Applying the distributive law to line 1, I obtain $(5 + 10^{k - 1} + 15^{k - 1}) - 6 - 12^{k -2} - 18^{k - 2} + 2$, and I cannot see how the second line is obtained from this. – Edward.Lin Jun 10 '21 at 11:36
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    @Edward.Lin $3\cdot2^5$ means $3\cdot2\cdot2\cdot2\cdot2\cdot2$. That is not the same as $6^5$ which is $6\cdot6\cdot6\cdot6\cdot6$. – Jaap Scherphuis Jun 10 '21 at 11:40
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    The distributive law means that the multiplier of a sum is applies to each term in the sum separately. So $$5\left(1+2^{k-1}+3^{k-1}\right)=5\cdot1+5\cdot2^{k-1}+5\cdot3^{k-1}.$$ It is not valid to try to combine the coefficient with the base of the exponent, as Jaap Scherphuis has pointed out. – Will Orrick Jun 10 '21 at 11:41
  • @Will Orrick: Understood now. So I need to learn the laws for how multiplicaton and exponentiation interact, from Kahn Academy (as suggested above). I don't think I ever learnt these at school. – Edward.Lin Jun 10 '21 at 11:56