數列問題及數學歸納法

1.設數列<an>，已知a1=2且滿足an+1=3an -2(n>=1)
試求數列an一般項的通式 (A:an=3^(n-1)+1)


a(1) = 2
a(n+1) = 3a(n)-2, n≧1

則當 n > 1,
a(n) = 3a(n-1) - 2
= 3{3a(n-2)-2} - 2
= 3{3[3a(n-3)-2]-2}-2
= 3^3 a(n-3) - 3^2．2 - 3．2 - 2
= 3^{n-1} a(1) - 2(3^{n-2}+3^{n-3}+...+3+1)
= 3^{n-1}．2 -2 (3^{n-1}-1}/2
= 3^{n-1} + 1

數學歸納法驗證:
a(1) = 3^0 + 1 = 2, true.

設 n = k 時, a(n) = a(k) = 3^{k-1}+1
則當 n = k+1 時,
a(n) = a(k+1) = 3a(k)-2 = 3(3^{k-1}+1)-2 = 3^k + 1 = 3^{n-1}+1, true.




2兩有理數列<an><bn>滿足關係式(2+√3)^n=an+bn√3，
利用數學歸納法證明:對任意正整數n，an，bn都是正整數

n=1 時. (2+√3)^1 = 2+(1)√3, a(1) = 2, b(1) = 1 均為正整數.

設 n = k 時, (2+√3)^k = a(k)+b(k)√n 其中 a(k), b(k) 均為正整數.
則當 n = k+1 時,
(2+√3)^n = (2+√3)(2+√3)^k = (2+√3)(a(k)+b(k)√3)
= 2a(k)+3b(k) +a(k)√3+2b(k)√3
= (2a(k)+3b(k)) + (a(k)+2b(k))√3
即 a(k+1) = 2a(k)+3b(k), b(k+1) = a(k)+2b(k)
正整數相乘, 相加, 其結果仍是正整數,
依假設 a(k), b(k) 均為正整數, 故 a(k+1), b(k+1) 亦均為正整數.

故得證.