Mad Man
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#1
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I get the jist of pbi but there are some cases where I think how on earth would I think that in the exam. IF you go to the link below in the last example, when n=k+2 how does the 3rd line simplify to the fourth line?

https://iitutor.com/mathematical-ind...2%88%88%20I%20.
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Sinnoh
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PBI often involves manipulating something from the assumed identity. A recurring theme in these questions is that when you're evaluating the case for n = k+1 or k+2, the expression for when n = k is 'hidden' in there, and the assumption you made about for n = k is required to build a case for n = k+1
Often when you're working with indices, this means reducing the index and bringing it down in front, e.g. 25(5k) instead of 5k+2
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Mad Man
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(Original post by Sinnoh)
PBI often involves manipulating something from the assumed identity. A recurring theme in these questions is that when you're evaluating the case for n = k+1 or k+2, the expression for when n = k is 'hidden' in there, and the assumption you made about for n = k is required to build a case for n = k+1
Often when you're working with indices, this means reducing the index and bringing it down in front, e.g. 25(5k) instead of 5k+2
But how does it simplify to the fourth line?
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Sinnoh
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(Original post by Mad Man)
But how does it simplify to the fourth line?
Do you mean going from 4^2 \times 4^k + 5^2 \times 5^k + 6^2 \times 6^k to 4^2 (15M - 5^k - 6^k) + 5^2 \times 5^k + 6^2 \times 6^k?
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Mad Man
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(Original post by Sinnoh)
Do you mean going from 4^2 \times 4^k + 5^2 \times 5^k + 6^2 \times 6^k to 4^2 (15M - 5^k - 6^k) + 5^2 \times 5^k + 6^2 \times 6^k?
No but it's okay, I understand now.
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Mad Man
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Why does, in 3) when you do (5+3) x 8^k-3 x 3^k that extra 3 disappear?
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RDKGames
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(Original post by Mad Man)
Why does, in 3) when you do (5+3) x 8^k-3 x 3^k that extra 3 disappear?
It does not disappear. Look at the term 3(8^k - 3^k). It is moved here.
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Mad Man
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(Original post by RDKGames)
It does not disappear. Look at the term 3(8^k - 3^k). It is moved here.
Shouldn't it be 3^k+1?
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RDKGames
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(Original post by Mad Man)
Shouldn't it be 3^k+1?
No, why?
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Mad Man
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#10
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(Original post by RDKGames)
No, why?
Can you explain to me what is going on here please? I'm stuck.
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RDKGames
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#11
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(Original post by Mad Man)
Can you explain to me what is going on here please? I'm stuck.
\underbrace{(5+3)\cdot 8^k}_{\text{expand}} - 3 \cdot 3^k

= 5 \cdot 8^k + \underbrace{3\cdot 8^k - 3 \cdot 3^k}_{\text{factorise}}

= 5 \cdot 8^k + 3(8^k - 3^k)
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Mad Man
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(Original post by RDKGames)
\underbrace{(5+3)\cdot 8^k}_{\text{expand}} - 3 \cdot 3^k

= 5 \cdot 8^k + \underbrace{3\cdot 8^k - 3 \cdot 3^k}_{\text{factorise}}

= 5 \cdot 8^k + 3(8^k - 3^k)
Oh that makes sense, I like that dot notation, it is easier to understand.

Also, suppose you have 15(15^k) - 3(3^k)
Can you do (15-3)(15^k-3^k)=12f(k)?
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RDKGames
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(Original post by Mad Man)
Oh that makes sense, I like that dot notation, it is easier to understand.

Also, suppose you have 15(15^k) - 3(3^k)
Can you do (15-3)(15^k-3^k)=12f(k)?
Nope.

15^{k+1} - 3^{k+1} = 3^{k+1} \cdot 5^{k+1} - 3^{k+1} = 3^{k+1}(5^{k+1} - 1)
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Mad Man
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#14
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(Original post by RDKGames)
Nope.

15^{k+1} - 3^{k+1} = 3^{k+1} \cdot 5^{k+1} - 3^{k+1} = 3^{k+1}(5^{k+1} - 1)
How can I get an A* in further maths, because every question I answer is slightly wrong
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DFranklin
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(Original post by Mad Man)
How can I get an A* in further maths, because every question I answer is slightly wrong
Practice. Should also be said that this is quite a tough question IMHO.

I'm not clear if questions like this come up in FM often enough for it to be worth learning it, but a basic knowledge of modular arithmetic really helps.

For example, in this case it quickly points you to the fact that 4^(2k+1) - 4 is always a multiple of 15, as is 5^(2k+1) - 5 and 6^(2k+1)-6; you can either prove this by induction (after which the actual result is trivial) or at least use it to guide you.
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