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FP1 June 2012 Q5

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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Zacken
22
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?
Reply 2
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emilyp2206
OP
2
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???
Reply 3
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Zacken
22
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=kn=k: that is: ∑r=1k4×3r=6(3k−1)\sum_{r=1}^k 4 \times 3^r = 6(3^k -1).

Now you want to show that ∑r=1k+14×3r=6(3k+1−1)\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: ∑r=1k+14×3r=∑r=1k(4×3r)+(4×3k+1)\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.
Reply 4
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emilyp2206
OP
2
Original post by Zacken
Assume the result holds for n=kn=k: that is: ∑r=1k4×3r=6(3k−1)\sum_{r=1}^k 4 \times 3^r = 6(3^k -1).

Now you want to show that ∑r=1k+14×3r=6(3k+1−1)\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: ∑r=1k+14×3r=∑r=1k(4×3r)+(4×3k+1)\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 :biggrin:D
Reply 5
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Zacken
22
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 :biggrin:D


Glad it's made sense. :biggrin:

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