各项均为正数的数列﹛an﹜,满足a1=1,a的n+1次方﹣a²=2(n属于正整数)①求数列﹛an﹜的通项公式②求数列﹛an²/2ⁿ﹜的前n项和Sn
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 13:54:13
![各项均为正数的数列﹛an﹜,满足a1=1,a的n+1次方﹣a²=2(n属于正整数)①求数列﹛an﹜的通项公式②求数列﹛an²/2ⁿ﹜的前n项和Sn](/uploads/image/z/2579292-36-2.jpg?t=%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%2C%E6%BB%A1%E8%B6%B3a1%3D1%2Ca%E7%9A%84n%2B1%E6%AC%A1%E6%96%B9%EF%B9%A3a%26%23178%3B%3D2%28n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E2%91%A0%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E2%91%A1%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9Ban%26%23178%3B%EF%BC%8F2%26%238319%3B%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
各项均为正数的数列﹛an﹜,满足a1=1,a的n+1次方﹣a²=2(n属于正整数)①求数列﹛an﹜的通项公式②求数列﹛an²/2ⁿ﹜的前n项和Sn
各项均为正数的数列﹛an﹜,满足a1=1,a的n+1次方﹣a²=2(n属于正整数)
①求数列﹛an﹜的通项公式②求数列﹛an²/2ⁿ﹜的前n项和Sn
各项均为正数的数列﹛an﹜,满足a1=1,a的n+1次方﹣a²=2(n属于正整数)①求数列﹛an﹜的通项公式②求数列﹛an²/2ⁿ﹜的前n项和Sn
a的n+1次方﹣a²=2(n属于正整数)是不是
【a(n+1)】² - 【an】² =
是的话那么
an²即为首项为a1²,公差为2的等差数列,
即an² = 1 + 2(n - 1)= 2n - 1
an =根号下 (2n - 1)
an²/2ⁿ = (2n -1)/2ⁿ = 2n / 2ⁿ - 1 / 2ⁿ
记bn = 2n / 2ⁿ Tn为bn的前n项和
Tn = 2 / 2 + (2×2)/(2²) + (2×3)/(2³)…… + 2n / 2ⁿ ……①
2Tn= 2 / 1 +(2×2)/ 2 +(2×3)/(2²)+……+ 2n/2^(n-1)……②
② - ①得
Tn = 2 + 2 / 2 + 2 / 2²+……+2/2^(n-1)- 2n / 2ⁿ
=2×(1 - (1/2)ⁿ)/(1 - 1/2) - 2n / 2ⁿ
=4 ×(1- 2^(-n )) - 2n × 2^(-n)
Sn=Tn -1/2 - (1/2)² - (1/2)³ - …… - (1 / 2)ⁿ
=Tn-(1/2)×(1 - (1/2)ⁿ)/(1 - 1/2)
=Tn- (1 - 2^(-n ))
=4 ×(1- 2^(-n )) - 2n × 2^(-n)- (1 - 2^(-n ))
=3 - 3 × 2^(-n ) - 2n × 2^(-n)
题目是不是错了啊?还是你的表述有问题?