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Proof by Induction

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Reply 20
Avatar for bong
bong
Yes ive got it now its quite a nice question once you complete it
Reply 21
Avatar for nuodai
nuodai
17
Original post by Mr M
I suggested expanding it all for a reason, students with less ability than you struggle with anything but the simplest factorisation.


I hadn't noticed you'd suggested that. The reason I wasn't keen to suggest expanding everything is that (as in raheem94's post) you get a lot of terms, and I'm not convinced that manipulating those would be easier than keeping it 'as factorized as possible' whilst doing the working.

I could perhaps have been clearer about the +1 not being included in the common factor, though. (But I did mention it in a previous post.)
Original post by nuodai
I hadn't noticed you'd suggested that. The reason I wasn't keen to suggest expanding everything is that (as in raheem94's post) you get a lot of terms, and I'm not convinced that manipulating those would be easier than keeping it 'as factorized as possible' whilst doing the working.


You only get 5 terms?!
(edited 11 years ago)
Reply 23
Avatar for raheem94
raheem94
13
Original post by nuodai
I hadn't noticed you'd suggested that. The reason I wasn't keen to suggest expanding everything is that (as in raheem94's post) you get a lot of terms, and I'm not convinced that manipulating those would be easier than keeping it 'as factorized as possible' whilst doing the working.

I could perhaps have been clearer about the +1 not being included in the common factor, though. (But I did mention it in a previous post.)


I like to look at the final answer i should get, and try to expand it, to see which steps i need to do to get the answer.
14(3k(2k−1)+1+4(k+1)3k)\frac{1}{4}(3^k (2k-1) + 1 + 4(k+1)3^k)

14(3k×2k−3k+1+4k×3k+4×3k)\frac{1}{4}(3^k \times 2k - 3^k + 1 + 4k \times 3^k + 4 \times 3^k)

14(3k×6k+1+3×3k)\frac{1}{4}(3^k \times 6k + 1 + 3 \times 3^k)

14(3k(6k+3)+1)\frac{1}{4}(3^k ( 6k + 3)+ 1)

and home.
Reply 25
Avatar for nuodai
nuodai
17
Original post by Mr M
You only get 5 terms?!


I think this must be a matter of style that varies from person to person. I find it much easier to make mistakes when I have a collection of terms that look similar (owing, for example, to having common factors) than I do by keeping the common factors out of the manipulation as much as possible. I suppose others might find it hard to look at an expression with lots of brackets and separate in their minds what stays the same and what doesn't.

Original post by Mr M
14(3k(2k−1)+1+4(k+1)3k)\frac{1}{4}(3^k (2k-1) + 1 + 4(k+1)3^k)

14(3k×2k−3k+1+4k×3k+4×3k)\frac{1}{4}(3^k \times 2k - 3^k + 1 + 4k \times 3^k + 4 \times 3^k)

14(3k×6k+1+3×3k)\frac{1}{4}(3^k \times 6k + 1 + 3 \times 3^k)

14(3k(6k+3)+1)\frac{1}{4}(3^k ( 6k + 3)+ 1)

and home.


14(3k(2k−1)+1+4(k+1)3k)\dfrac{1}{4}(3^k (2k-1) + 1 + 4(k+1)3^k)

=14(3k(2k−1+4(k+1))+1)=\dfrac{1}{4}( 3^k(2k-1+4(k+1)) + 1)

=14(3k(2k−1+4k+4)+1)=\dfrac{1}{4}( 3^k(2k-1+4k+4) + 1)

=14(3k(6k+3)+1)=\dfrac{1}{4}( 3^k(6k+3) + 1)

Like I say, matter of style.
(edited 11 years ago)
Original post by nuodai
I think this must be a matter of style that varies from person to person. I find it much easier to make mistakes when I have a collection of terms that look similar (owing, for example, to having common factors) than I do by keeping the common factors out of the manipulation as much as possible. I suppose others might find it hard to look at an expression with lots of brackets and separate in their minds what stays the same and what doesn't.


Common factors is by far the best method. Weaker students are unable to do it though but they often can experience success by expanding, collecting like terms and then factorising.
Reply 27
Avatar for nuodai
nuodai
17
Original post by Mr M
Common factors is by far the best method. Weaker students are unable to do it though but they often can experience success by expanding, collecting like terms and then factorising.


I think I'll have to bow to your superior experience on this one. I suppose it can't hurt for the OP to see two equivalent methods.

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