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    can anyone help me out with the method for this question?
    Proove by induction for all positive integers of n;
    (sum from r=1 to n) 4x3r = 6(3n - 1)
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    (Original post by emilyp2206)
    can anyone help me out with the method for this question?
    Proove by induction for all positive integers of n;
    (sum from r=1 to n) 4x3r = 6(3n - 1)
    What have you done so far?
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    (Original post by Zacken)
    What have you done so far?

    prooved for n=1 but to be honest that could be winged, im unfamilar with how to deal with the power of r, do i sub in k? do i sub in k+1???
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    (Original post by emilyp2206)
    prooved for n=1 but to be honest that could be winged, im unfamilar with how to deal with the power of r, do i sub in k? do i sub in k+1???
    Assume the result holds for n=k: that is: \sum_{r=1}^k 4 \times 3^r = 6(3^k -1).

    Now you want to show that \displaystyle \sum_{r=1}^{k+1} 4 \times 3^r = 6(3^{k+1} - 1).

    To do so, you'll need to split the sum as such: \displaystyle \sum_{r=1}^{k+1} 4\times 3^r = \sum_{r=1}^k (4\times 3^r) + (4\times 3^{k+1}).

    You assumed something about the first term/sum in that equality, substitute that assumption in and work towards the answer.
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    (Original post by Zacken)
    Assume the result holds for n=k: that is: \sum_{r=1}^k 4 \times 3^r = 6(3^k -1).

    Now you want to show that \displaystyle \sum_{r=1}^{k+1} 4 \times 3^r = 6(3^{k+1} - 1).

    To do so, you'll need to split the sum as such: \displaystyle \sum_{r=1}^{k+1} 4\times 3^r = \sum_{r=1}^k (4\times 3^r) + (4\times 3^{k+1}).

    You assumed something about the first term/sum in that equality, substitute that assumption in and work towards the answer.
    Thank you for the help!! think now ive chilled after the dreadful core one its looking abit clearer!

    worked it through and factorized to get the answer D
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    (Original post by emilyp2206)
    Thank you for the help!! think now ive chilled after the dreadful core one its looking abit clearer!

    worked it through and factorized to get the answer D
    Glad it's made sense.
 
 
 
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