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Proof by induction divisibility question

need help how you get from one part to the other see images
Ok so i've got to the part where you do f(k+1)-f(k) but now stuck as to how you get to 3(fk)+21x5^2k-1
proof by induction q.PNG4.3KB
proof by induction ans.PNG73.3KB
Reply 2
Avatar for mqb2766
mqb2766
21
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)
Original post by Yr_11_MATHS
Ok so i've got to the part where you do f(k+1)-f(k) but now stuck as to how you get to 3(fk)+21x5^2k-1
Original post by mqb2766
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)


sorry?
Is this edexcel a level maths
Original post by Yr_11_MATHS
Ok so i've got to the part where you do f(k+1)-f(k) but now stuck as to how you get to 3(fk)+21x5^2k-1
let me send pic of what ive currently got
Original post by mqb2766
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)
DSC_0002.JPG283.9KB
Original post by BonsaiColony
Is this edexcel a level maths


yes edexcel further
Reply 7
Avatar for mqb2766
mqb2766
21
4*4^(k+1) - 4^(k+1) + 25*5^(2k-1) - 5^(2k-1)
Rearrange
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)
Gives
3*f(k) + 21*5^(2k-1)

Original post by mqb2766
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)
Original post by mqb2766
4*4^(k+1) - 4^(k+1) + 25*5^(2k-1) - 5^(2k-1)
Rearrange
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)
Gives
3*f(k) + 21*5^(2k-1)

Ok so you basically collected like terms and then arrange to find a multiple of the f(k) (in this question being 3 because there will be no (4^k+1) remainder thus leaving to a part which will leave you with a remainder of 21(5^2k-1)....

Thank you for your help bro :smile:
Original post by mqb2766
4*4^(k+1) - 4^(k+1) + 25*5^(2k-1) - 5^(2k-1)
Rearrange
4^(k+1)*(4-1) + 3*5^(2k-1) + 5^(2k-1)*(22 - 1)
Gives
3*f(k) + 21*5^(2k-1)


for some reason it's not letting me rep you cos i need to "spread it around"

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