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    http://files.physicsandmathstutor.co...%20Edexcel.pdf
    n=1
    \begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}

    \therefore true\ for\ n=1


    Assume true for n=k that

    \begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}

    what am i supposed to do next? it doesn't give me something to substitute something somewhere?
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    (Original post by huiop)
    http://files.physicsandmathstutor.co...%20Edexcel.pdf
    n=1
    \begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}

    \therefore \true \for n=1


    Assume true for n=k that

    \begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}

    what am i supposed to do next? it doesn't give me something to substitute something somewhere?
    I might be wrong, but I have a feeling the kth term is the determinant...
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    (Original post by huiop)
    http://files.physicsandmathstutor.co...%20Edexcel.pdf
    n=1
    \begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}

    \therefore true\ for\ n=1


    Assume true for n=k that

    \begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}

    what am i supposed to do next? it doesn't give me something to substitute something somewhere?
    But can you not do it for K+1, and use the result from the LHS of the equation in your assumption as a substitute?
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    (Original post by huiop)
    http://files.physicsandmathstutor.co...%20Edexcel.pdf
    n=1
    \begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}

    \therefore true\ for\ n=1


    Assume true for n=k that

    \begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}

    what am i supposed to do next? it doesn't give me something to substitute something somewhere?
    Test for n=1
    Assume true for n=k.
    Prove for n=k+1. This would require you you to post-multiply both sides by the matrix 1, -1, 0, 5. (You know which one, cba to post it in the form) and factor every element to show it works for n=k+1.
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    (Original post by huiop)
    http://files.physicsandmathstutor.co...%20Edexcel.pdf
    n=1
    \begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}

    \therefore true\ for\ n=1


    Assume true for n=k that

    \begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}

    what am i supposed to do next? it doesn't give me something to substitute something somewhere?
    Consider the matrix to the power of k + 1,and split that into  \begin {pmatrix} 1&0 \\-1&5 \end {pmatrix}^{k}  \begin {pmatrix} 1&0 \\-1&5 \end {pmatrix} and sub that kth power matrix for the general formula we assumed was true for n = k.
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    (Original post by huiop)
    http://files.physicsandmathstutor.co...%20Edexcel.pdf
    n=1
    \begin {pmatrix}1&0\\-\frac{1}{4}(5^1 -1)&5^1\end {pmatrix}

    \therefore true\ for\ n=1


    Assume true for n=k that

    \begin {pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k \end {pmatrix}

    what am i supposed to do next? it doesn't give me something to substitute something somewhere?
    Consider multiplying the RHS by \begin {pmatrix}1&0\\-1&5 \end {pmatrix} on the right. Can you see why?
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    (Original post by Cadherin)
    I might be wrong, but I have a feeling the kth term is the determinant...
    Not needed for this proof. Though the determinant is 5^n for all n\geq 1 ; n\in \mathbb{Z_+}
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    (Original post by Slowbro93)
    But can you not do it for K+1, and use the result from the LHS of the equation in your assumption as a substitute?
    (Original post by RDKGames)
    Test for n=1
    Assume true for n=k.
    Prove for n=k+1. This would require you you to post-multiply both sides by the matrix 1, -1, 0, 5. (You know which one, cba to post it in the form) and factor every element to show it works for n=k+1.
    u sure i can do that? i considered it but i wasn't too sure if i was allowed to do that since it didn't state that matrix without to the power of n on it
    (Original post by AMarques)
    Consider the matrix to the power of k + 1,and split that into  \begin {pmatrix} 1&0 \\-1&5 \end {pmatrix}^{k}  \begin {pmatrix} 1&0 \\-1&5 \end {pmatrix} and sub that kth power matrix for the general formula we assumed was true for n = k.

    (Original post by Paraphilos)
    Consider multiplying the RHS by \begin {pmatrix}1&0\\-1&5 \end {pmatrix} on the right. Can you see why?
    so you get the identity matrix on the LHS?
    Edit:^^ no u don't but u get 1 of tht matrix multiplied by the kth matrix which is to the power k+1 right?
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    (Original post by huiop)
    u sure i can do that? i considered it but i wasn't too sure if i was allowed to do that since it didn't state that matrix without to the power of n on it
    Of course. Just like with summation of series when you add the (k+1)th term onto both sides, you multiply here by the same matrix as you're working with exponents.

    Let A=\begin {pmatrix}1 & 0 \\ {-1} & {5} \end {pmatrix}

    Then it follows that A^{k+1}=(A^k)(A^1) (think of it as performing a transformation A a total amount of k times, then doing an additional one)

    And you know what A^k is because you assumed it to be true for n=k
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    (Original post by RDKGames)
    Of course. Just like with summation of series when you add the (k+1)th term onto both sides, you multiply here by the same matrix as you're working with exponents.

    Let A=\begin {pmatrix}1 & 0 \\ {-1} & {5} \end {pmatrix}

    Then it follows that A^{k+1}=(A^k)(A^1) (think of it as performing a transformation A a total amount of k times, then doing an additional one)

    And you know what A^k is because you assumed it to be true for n=k
    i've got

    \begin {pmatrix}1&0\\{[-\frac{1}{4}(5^k -1)]-5^k}&5^k \times 5 \end {pmatrix}

    the 5^k \times 5=5^{k+1}

    but the other meaty term i've only gotten this far

    -\frac{1}{4}[5^k -1+(5^k \times 4)]
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    (Original post by huiop)
    i've got

    \begin {pmatrix}1&0\\{[-\frac{1}{4}(5^k -1)]-5^k}&5^k \times 5 \end {pmatrix}

    the 5^k \times 5=5^{k+1}

    but the other meaty term i've only gotten this far

    -\frac{1}{4}[5^k -1+(5^k \times 4)]
    Is it really meaty though? 5^k+4\cdot 5^k=5\cdot 5^k = 5^{k+1}
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    (Original post by RDKGames)
    Is it really meaty though? 5^k+4\cdot 5^k=5\cdot 5^k = 5^{k+1}
    Looks like we have another student (in the form of Mr or Mrs Huiop) who doesn't know their laws of indices. Typical.
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    (Original post by HapaxOromenon3)
    Looks like we have another student (in the form of Mr or Mrs Huiop) who doesn't know their laws of indices. Typical.
    Those are common around here.
 
 
 
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