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
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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
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
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
(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
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...
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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
收起
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...
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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
收起
2^(n+1)+n-2