Question

已知(b1-1)/2+(b2-1)/(2的平方)+(b3-1)/2的3次方+.......加到(bn-1)/(2的N次方)=n,求数列bn的前N项和

Answer 1

(b1-1)/2+(b2-1)/(2的平方)+.......加到(bn-1)/(2的N次方)=n

(b1-1)/2+(b2-1)/(2的平方)+.......加到(b（n-1）-1)/(2的N-1次方)=n-1

两式左右相减

(bn-1)/(2的N次方)=1

bn＝1+2^n

Sn=b1+……+bn=n+2^(n+1)-2=(n-2)+2^(n+1)

2^n即2的n次方

Answer 2

n
(b1-1)/2+(b2-1)/(2的平方)+(b3-1)/2的3次方+.......加到(bn-1)/(2的N次方)=n
n+1
(b1-1)/2+(b2-1)/(2的平方)+(b3-1)/2的3次方+.......加到(bn+1-1)/(2的N+1次方)=n+1

两式相减
(bn+1-1)/2^(n+1)=1
=>
bn+1=2^(n+1)+1
当n>=2时
bn=2^n+1
当n=1时
(b1-1)/2=1
b1=2=2^1+1满足上式
=>
bn的前N项和
=(2+2^2+……+2^n)+n
=2*(2^n-1)/(2-1)+n
=2^(n+1)+n-2

Answer 3

我来回答！！这题目很难打字！！你得保证分数给我才可以，呵呵
(b1-1)/2+(b2-1)/(2的平方)+.......加到(bn-1)/(2的n次方)=n
用n-1代替得到

(b1-1)/2+(b2-1)/(2的平方)+.......加到(b（n-1）-1)/(2的n-1次方)=n-1

两式相减

(bn-1)/(2的n次方)=1

bn＝2^n +1
2^n市等比数列！！用钱n相合公式得到

Sn=b1+……+bn=n+2^(n+1)-2

Answer 4

2^(n+1)+n-2