S=1+(1+2)+(1+2+2^2)+ … +(1+2+2^2+…+2^10)的值
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![S=1+(1+2)+(1+2+2^2)+ … +(1+2+2^2+…+2^10)的值](/uploads/image/z/3976353-9-3.jpg?t=S%3D1%2B%281%2B2%29%2B%281%2B2%2B2%5E2%29%2B+%E2%80%A6+%2B%281%2B2%2B2%5E2%2B%E2%80%A6%2B2%5E10%29%E7%9A%84%E5%80%BC)
S=1+(1+2)+(1+2+2^2)+ … +(1+2+2^2+…+2^10)的值
S=1+(1+2)+(1+2+2^2)+ … +(1+2+2^2+…+2^10)的值
S=1+(1+2)+(1+2+2^2)+ … +(1+2+2^2+…+2^10)的值
等比求和知1+2+2^2+… +2^n=2^n-1
所以
1=2^1-1
1+2=2^2-1
1+2+2^2=2^3-1
… … … …
1+2+2^2+…+2^10=2^10-1
所以S=(2^1-1)+(2^2-1)+(2^3-1)+…+(2^10-1)
=(2^1+2^2+2^3+…+2^10)+(-1)*10
=2(2^10-1)/(2-1)-10
=2036
数列的通项公式an=a1q^(n-1)=1*2^(n-1)=2^(n-1)
S=a1+a2+...+an
=2^0+2^1+2^3+...+2^(n-1)
=1*[1-2^(n-1)]/(1-2)
=2^(n-1)-1