MATH INDUCTION

2014-02-04 5:36 pm

1)Using induction, prove that 3^(2n+3)+40n-27 is divisible by 64 for all integer n≥1

Clearly indicate the base case and the induction hypothesis.

What I did...

n=1

(3^(5)+40-27)/64 = 4 True that 3^(2n+3)+40n-27 is divisible by 64 for all integer n≥1

Assume
3^(2k+3)+40k-27 is divisible by 64
Show
3^(2(k+1)+3)+40(k+1)-27 is divisible by 64

(3^(2(k+1))+40(k+1)-27)/64 = (3^(2k+3)+40k-27)/64 + the statement n term.

Here is my problem, what is the n term?

the statement that I found is
4,35,309,2770.....

I didn't see any relationship between them

Help!!

Thanks

回答 (1)

Ka ho

2014-02-04 8:21 pm

✔ 最佳答案

For n=1,
3^(2n+3)+40n-27=3^(5)+40-27=256=4(64)
∴The statement is true for n=1.

Assume the statement is true for n=k
i.e. 3^(2k+3)+40k-27=64M , where n,M are positive integers.

For n=k+1,
3^(2(k+1)+3)+40(k+1)-27
=3^(2k+3+2)+40k+13
=9(3^(2k+3))+40k+13
=9[64M+27-40k]+40k+13 {Note that 3^(2k+3)=64M+27-40k}
=576M+243-360k+40k+13
=576M-320k+256
=64(9M-5k+4)
∵9M-5k+4 is an integer
∴The statement is true for n=k+1.

By..........
-----------------------------------------
This is the correct approach to solve mathematical induction of division problem
It aims at proving that the term of n=k+1 can satisfy the statement by showing
it is one of the multiple of 64.

Your approach can only work on series / summation problems, please notice
that there is no any addition required in the statement. Eventually there won't be any "+" in the hypothesis.

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