在正项数列an中Sn=1/(根号a1+根号a2)+1/(根号a2+根号a3)+...+1/(根号an+根号an+1)(1)若an是首项为25,公差为2的等差数列,求S100(2)若Sn=np/(根号a1+根号an+1)(p是正常数)对正整数n恒成立,求证:an是等差数
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![在正项数列an中Sn=1/(根号a1+根号a2)+1/(根号a2+根号a3)+...+1/(根号an+根号an+1)(1)若an是首项为25,公差为2的等差数列,求S100(2)若Sn=np/(根号a1+根号an+1)(p是正常数)对正整数n恒成立,求证:an是等差数](/uploads/image/z/1775545-25-5.jpg?t=%E5%9C%A8%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97an%E4%B8%ADSn%3D1%2F%28%E6%A0%B9%E5%8F%B7a1%2B%E6%A0%B9%E5%8F%B7a2%29%2B1%2F%28%E6%A0%B9%E5%8F%B7a2%2B%E6%A0%B9%E5%8F%B7a3%29%2B...%2B1%2F%28%E6%A0%B9%E5%8F%B7an%2B%E6%A0%B9%E5%8F%B7an%2B1%29%281%29%E8%8B%A5an%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BA25%2C%E5%85%AC%E5%B7%AE%E4%B8%BA2%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82S100%282%29%E8%8B%A5Sn%3Dnp%2F%28%E6%A0%B9%E5%8F%B7a1%2B%E6%A0%B9%E5%8F%B7an%2B1%29%EF%BC%88p%E6%98%AF%E6%AD%A3%E5%B8%B8%E6%95%B0%EF%BC%89%E5%AF%B9%E6%AD%A3%E6%95%B4%E6%95%B0n%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E8%AF%81%EF%BC%9Aan%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0)
在正项数列an中Sn=1/(根号a1+根号a2)+1/(根号a2+根号a3)+...+1/(根号an+根号an+1)(1)若an是首项为25,公差为2的等差数列,求S100(2)若Sn=np/(根号a1+根号an+1)(p是正常数)对正整数n恒成立,求证:an是等差数
在正项数列an中Sn=1/(根号a1+根号a2)+1/(根号a2+根号a3)+...+1/(根号an+根号an+1)
(1)若an是首项为25,公差为2的等差数列,求S100
(2)若Sn=np/(根号a1+根号an+1)(p是正常数)对正整数n恒成立,求证:an是等差数列
只要第二小题的过程就行
在正项数列an中Sn=1/(根号a1+根号a2)+1/(根号a2+根号a3)+...+1/(根号an+根号an+1)(1)若an是首项为25,公差为2的等差数列,求S100(2)若Sn=np/(根号a1+根号an+1)(p是正常数)对正整数n恒成立,求证:an是等差数
第二小题若无头绪以数学归纳法入手
若Sn=np/(√a1+√a(n+1))(p是正常数)对正整数n恒成立,所以对S1也恒成立
1/(√a1+√a2) = 1*p/(√a1+√a2) 所以p等于1,不妨设a2 =a1+d
数学归纳法 证明数列an 公差为d,即 an = a(n-1) +d
1# k=1 时,显然 a2=a1+d
2# 若 k= n-1时 an = a(n-1)+d
3# 当 k =n 时
Sn=S(n-1)+1/(√an+√a(n+1)) = (n-1)/(√a1+√an) + 1/(√an+√a(n+1)) = n/(√a1+√a(n+1))
所以 (n-1)[1/(√a1+√a(n+1))-1/(√a1+√an) ]= 1/(√an+√a(n+1)) - /(√a1+√a(n+1))
即 (n-1)[√an-√a(n+1)]/[(√a1+√a(n+1)*(√a1+√an) ] = (√a1 - √an)/[(√a1+√a(n+1))*(√an+√a(n+1)) ]
(n-1)[√an-√a(n+1)]/(√a1+√an) = (√a1 - √an)/[(√an+√a(n+1)]
所以 (n-1)[√an-√a(n+1)][(√an+√a(n+1)] = (√a1 - √an)(√a1+√an)
(n-1)(a(n+1)-an) = (an -a1) = (n-1)d
所以 an+1 - an =d
所以,命题an = a(n-1) +d(n∈Z+) 成立,an 是等差数列