早教吧作业答案频道 -->数学-->
求数列{n^2*2^n}的前n项和(高二数列问题)注意啊是n的平方再乘以2的n次方啊
题目详情
求数列{n^2*2^n}的前n项和(高二数列问题)
注意啊 是n的平方再乘以2的n次方啊
注意啊 是n的平方再乘以2的n次方啊
▼优质解答
答案和解析
a(n) = n^2*2^n = n(n+1)2^n - n2^n = b(n) - c(n).
C(n) = c(1)+c(2)+c(3) + ... + c(n-1) + c(n)
= 1*2 + 2*2^2 + 3*2^3 + ... + (n-1)2^(n-1) + n2^n
2C(n) = 1*2^2 + 2*2^3 + ... + (n-1)2^n + n2^(n+1),
C(n) = 2C(n) - C(n) = - 2 - 2^2 - 2^3 - ... - 2^n + n2^(n+1)
= n2^(n+1) - 2[1 + 2 + ... + 2^(n-1)]
= n2^(n+1) - 2[2^n - 1]/(2-1)
= n2^(n+1) - 2^(n+1) + 2
= 2 + (n-1)2^(n+1).
B(n) = b(1)+b(2)+b(3) + ... + b(n-1) + b(n)
= 1*2*2^1 + 2*3*2^2 + 3*4*2^3 + ... + (n-1)n2^(n-1) + n(n+1)2^n
2B(n) = 1*2*2^2 + 2*3*2^3 + ... + (n-1)n2^n + n(n+1)2^(n+1),
B(n) = 2B(n) - B(n) = -2*1*2^1 - 2*2*2^2 - 2*3*2^3 - ... - 2*n*2^n + n(n+1)2^(n+1)
= n(n+1)2^(n+1) - 2[1*2^1 + 2*2^2 + 3*2^3 + ... + n*2^n]
= n(n+1)2^(n+1) - 2C(n)
s(n) = a(1)+a(2)+...+a(n) = b(1)+b(2)+...+b(n) - c(1) - c(2) - ... - c(n) = B(n) - C(n)
= n(n+1)2^(n+1) - 2C(n) - C(n)
= n(n+1)2^(n+1) - 3C(n)
= n(n+1)2^(n+1) - 3[2 + (n-1)2^(n+1)]
= n(n+1)2^(n+1) - 6 - 3(n-1)2^(n+1)
= [n^2 + n - 3n + 3]2^(n+1) - 6
= [n^2 - 2n+3]2^(n+1) - 6
C(n) = c(1)+c(2)+c(3) + ... + c(n-1) + c(n)
= 1*2 + 2*2^2 + 3*2^3 + ... + (n-1)2^(n-1) + n2^n
2C(n) = 1*2^2 + 2*2^3 + ... + (n-1)2^n + n2^(n+1),
C(n) = 2C(n) - C(n) = - 2 - 2^2 - 2^3 - ... - 2^n + n2^(n+1)
= n2^(n+1) - 2[1 + 2 + ... + 2^(n-1)]
= n2^(n+1) - 2[2^n - 1]/(2-1)
= n2^(n+1) - 2^(n+1) + 2
= 2 + (n-1)2^(n+1).
B(n) = b(1)+b(2)+b(3) + ... + b(n-1) + b(n)
= 1*2*2^1 + 2*3*2^2 + 3*4*2^3 + ... + (n-1)n2^(n-1) + n(n+1)2^n
2B(n) = 1*2*2^2 + 2*3*2^3 + ... + (n-1)n2^n + n(n+1)2^(n+1),
B(n) = 2B(n) - B(n) = -2*1*2^1 - 2*2*2^2 - 2*3*2^3 - ... - 2*n*2^n + n(n+1)2^(n+1)
= n(n+1)2^(n+1) - 2[1*2^1 + 2*2^2 + 3*2^3 + ... + n*2^n]
= n(n+1)2^(n+1) - 2C(n)
s(n) = a(1)+a(2)+...+a(n) = b(1)+b(2)+...+b(n) - c(1) - c(2) - ... - c(n) = B(n) - C(n)
= n(n+1)2^(n+1) - 2C(n) - C(n)
= n(n+1)2^(n+1) - 3C(n)
= n(n+1)2^(n+1) - 3[2 + (n-1)2^(n+1)]
= n(n+1)2^(n+1) - 6 - 3(n-1)2^(n+1)
= [n^2 + n - 3n + 3]2^(n+1) - 6
= [n^2 - 2n+3]2^(n+1) - 6
看了 求数列{n^2*2^n}的前...的网友还看了以下:
数列{an}的前n项和记注意Sn ,a1=1,a(n+1)=(n+2)Sn/n(n=1,2,3`` 2020-04-06 …
已知等比数列an中,a1=2,前n项和为Sn,数列a(n+1)也是等比数列,求数列an的通项公式a 2020-05-13 …
高2文科北师大版必修5等比数列与等差数列综合题目求数列1,3a,5a²,7a³……(2n-1)[a 2020-05-14 …
数列(a n)的前N项和为Sn,满足点(an,Sn)在直线y=2X+1上.1.求数列(an)的通项 2020-05-15 …
/*2.【问题描述】 输入2 个正整数m 和n(m≥1,n≤500),统计并输出m 和n 之间的素 2020-05-17 …
初二课文——《小石潭记》~两个字的注音问题~“心乐之”的“乐”,有些辅导书上写的是lè,有的写yù 2020-06-16 …
数列{an}中,an=[(-1)^(n+1)]*(2n+1)求数列前n项和Sn注意用并项求和法~. 2020-07-11 …
绘制复式条形统计图,必须要写上标题和标注图例,直条宽度要一致,间隔要相等。[] 2020-11-07 …
==困惑的一题.(内含5分)若n⊥α,n⊥β,则α//β若m//n,m//α,则n//α注:α、β为 2020-12-02 …
奥运会一天天向我们大步走来,“如何与奥运冠军一同成长”已成为全国青少年的热门话题和关注的焦点。你班最 2020-12-21 …