數學歸納法證明

(1)
(4^n)-1可以被3整除

For n=1,
(4^1) - 1 = 3 ... which is divisible by 3
∴The proposition is true for n=1.

Next, assume the propostion is true for some positive integers k,
that is, (4^k)-1=3m, where m is a constant.

For n=k+1,
[ 4^(k+1) ] - 1
= 4‧4^(k) - 1
= 4‧(3m+1) - 1 ... [ by (4^k)-1 = 3k ]
= 3‧4m + 4 - 1
= 3‧4m + 3
= 3(4m + 1) ... which is divisible by 3

∴The propostion is true for n=k+1.
By the principle of mathematical induction, the proposition is true for all positive integer n.

(2)
對正整數n>4,(n+1)!>2^n+3

For n=5,
L.H.S.= (5+1)! = 6! = 720
R.H.S. = 2^5+3 = 35
L.H.S. > R.H.S.
∴The proposition is true for n=5.

Next, assume the proposition is true for some positive integers k,
that is, (k+1)!>2^k+3

For n=k+1,
L.H.S.
= (k+1+1)!
= (k+2)!
= (k+2)(k+1)!
> (k+2)(2^k+3) ... [ by (k+1)!>2^k+3 ]
= k(2^k+3) + 2‧(2^k+3)
= k(2^k+3) + 2‧2^k + 3 + 3
= k(2^k+3) + 3 + 2‧2^k + 3

R.H.S.
= 2^(k+1)+3
= 2‧2^k+3

∵k is a integer and k>4
∴ k(2^k+3) + 3 >0
∴L.H.S.>R.H.S
∴The propostion is true for n=k+1.
By the principle of mathematical induction, the proposition is true for all positive integer n.

2014-03-20 23:35:39 補充：
第1題第4行及第5行的propostion應是proposition

2014-03-21 00:48:33 補充：
第1題第6行where m is a constant.改做where m is positive integer會適合D

2014-03-21 18:28:44 補充：
唔係丫fa
我見佢第1題(4^n)-1既次方識用括號
咁2^n+3應該淨係講2嘅n次方 ...
| '.' |

(2) 的意思是 (n+1)! > 2^(n+3) ?

4^k - 1 = 3M
4^(k+1) - 1 = 4 × 4^k - 1 = 4 × (3M + 1) - 1 = 12 M + 3 = 3(4M + 1)

2014-03-20 23:05:22 補充：
(k+1)! > 2^k + 3
(k+2)! = (k+2)(k+1)! > (k+2)(2^k + 3) = k×2^k + 2^(k+1) + 3k + 6 > 2^(k+1) + 3

2014-03-22 21:53:15 補充：
其實 ☂雨後晴空☀ 講得有道理的，因為如果當真是

(n+1)! > 2^n + 3

那麼當 n > 2 就可以了，不必 n > 4。

不過發問者都已經一走了之～

HK~ 的答案也很有參考價值～