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

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