F.4 Mathematical Induction [復制鏈接]

Prove by mathematical induction that 8^n+(2*7^n)-1 is divisible by 7 for all positive integers n.

Prove by mathematical induction that (6^n+2)+(7^2n+1)is divisible by 43 for all positive integers n.

跟格式做
when n = 1
8+14-1 = 21 = 7(3)
S1 is true

assume Sk is true where k>=1
8^k + (2*7^k)-1 = 7M where M is an integer
8^k = 7M+1 - (2*7^k)

when n = k+1
8^k+1 +(2*7^k+1)-1
= 8*8^k + (14*7^k )- 1
= 8*[7M+1 - (2*7^k)] + (14*7^k )- 1
= 56M + 8 - 16*7^k + 14*7^k - 1
= 56M + 7 - 2*7^k
=7( 8M+1-2*7^(k-1))
=7N since 8M+1-2*7^(k-1) must be a integer

其實MI呢D...算係咁多課入面最易handle果幾課ga啦
操多d..CE呢一課係搶分ga..加油加油

Prove by mathematical induction that (6^n+2)+(7^2n+1)is divisible by 43 for all positive integers n.


呢題n=1都唔得

引用第3樓eric_ah4於2007-09-13 20:30發表的“”:
Prove by mathematical induction that (6^n+2)+(7^2n+1)is divisible by 43 for all positive integers n.
呢題n=1都唔得

When n=1,

(6^n+2)+(7^2n+1)
= 6^3 + 7^3
= 559
= 43 * 13

Why not?

引用第2樓樂仔於2007-09-13 19:35發表的“”:
其實MI呢D...算係咁多課入面最易handle果幾課ga啦
操多d..CE呢一課係搶分ga..加油加油

保合格既必做題目

引用第4樓kylau於2007-09-13 20:41發表的“”:
When n=1,
(6^n+2)+(7^2n+1)
.......

我以為係(6^1)+2 and (7^2n)+1

Prove by mathematical induction that (6^n+2)+(7^2n+1)is divisible by 43 for all positive integers n.

For n=1
LHS=6^3+7^3=559,which is divisible by 43

Assume it is true for n=k
ie. (6^k+2)+(7^2k+1)=43N

For n=k+1
LHS
=(6^k+3)+(7^2k+3)
=6(6^k+2)+(7^2k+3)
=6[(6^k+2)+(7^2k+1)]+(7^2k+3)-6(7^2k+1)
=258N+(7^2k+1)(49-6)
=43(6N+7^2k+1)

thx!

引用第8樓dinlo於2007-09-13 22:02發表的“”:
thx!

u use which textbook