题目内容
((本小题满分14分)
在数列,中,a1=2,b1=4,且成等差数列,成等比数列()
(Ⅰ)求a2,a3,a4及b2,b3,b4,由此猜测,的通项公式,并证明你的结论;
(Ⅱ)证明:.
在数列,中,a1=2,b1=4,且成等差数列,成等比数列()
(Ⅰ)求a2,a3,a4及b2,b3,b4,由此猜测,的通项公式,并证明你的结论;
(Ⅱ)证明:.
由条件得
由此可得
.································ 2分
猜测.································································ 4分
用数学归纳法证明:
①当n=1时,由上可得结论成立.
②假设当n=k时,结论成立,即
,
那么当n=k+1时,
.
所以当n=k+1时,结论也成立.
由①②,可知对一切正整数都成立.······························· 7分
(Ⅱ).
n≥2时,由(Ⅰ)知.·································· 9分
故
综上,原不等式成立. ············································································ 14分
由此可得
.································ 2分
猜测.································································ 4分
用数学归纳法证明:
①当n=1时,由上可得结论成立.
②假设当n=k时,结论成立,即
,
那么当n=k+1时,
.
所以当n=k+1时,结论也成立.
由①②,可知对一切正整数都成立.······························· 7分
(Ⅱ).
n≥2时,由(Ⅰ)知.·································· 9分
故
综上,原不等式成立. ············································································ 14分
略
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