Induction Q (1 Viewer)

[STx]

Q: If U_n=9ⁿ⁺¹-8n-9, show that U_n+1=9U_n+64n+64, and hence show that U_n is divisible by 64 for n≥1. I can get the first part, not quite the second one. thx

 

[onebytwo]

Q: If U_n=9ⁿ⁺¹-8n-9, show that U_n+1=9U_n+64n+64, and hence show that U_n is divisible by 64 for n≥1. I can get the first part, not quite the second one. thx

assume the stat. is true for n=k
Uk = 9^k+1 -8k-9 = 64P , to use this make 9^k+1 the subject
then, U(k+1) = 9^k+2 -8(k+1)-9
write this as 9.9^k+1-8(k+1)-9
by assumption and simplifying, we get = 9(Uk) + 64k+64
then = 9(64P) +64k + 64
then factorise 64 out, so its divisible

Hence by theory of MI, if its true for .....................

 

[STx]

another Q: If U_n=5²ⁿ+3n-1, show that U_n is divisible by 9 for n≥1.

 

[acmilan]

U_n = 5²ⁿ+3n-1 divisible by 9

U₁ = 5² + 3 - 1 = 27 = 3x9, so its true for n = 1

Assume its true for some integer n≥1
ie. U_n = 5²ⁿ+3n-1 = 9m for some integer m

then
U_n+1
= 5²⁽ⁿ⁺¹⁾+3(n+1)-1
= 5²ⁿ.5² + 3n + 3 - 1
= (9m - 3n + 1).25 + 3n + 2 (using the assumption)
= 25.9m - 75n + 25 + 3n + 2
= 25.9m - 72n + 27
= 9(25m - 8n + 3)
= 9k for some integer k

Hence U_n+1 is divisible by 9 and the result follows by induction

 

[STx]

ah yes, damn i just did it as well, thx anyway acmilan