題目有誤,請(qǐng)改正

抱歉題目前一部分改正為：

對(duì)于n∈N^*,將n表示為n=a₀×2^^k+a₁×2^^k-1+a₂×2 ^^k-2 +…+a_k-1×2^ ¹ +a_k×2 ^⁰;當(dāng)...

謝謝。 

把正整數(shù)依次表示為二進(jìn)制數(shù)：
1,
10, 11,
100,101,110,111;
1000,1001,1010，1011,1100，1101,1110， 1111;
10000,10001,10010,10011,10100,10101,10110,10111,

...n...1...2...3...4...5...6...7...8...9...10...11...12...13...14...15...
I(n)...0...1...0...2...1...1...0...3...2...2......1......2.....1.....1....0....
...n...16...17...18...19...20...21...22...23...
I(n)....4....3......3.....2.....3.....2.....2.....1
在二進(jìn)制中，一位數(shù)1個(gè)，兩位數(shù)2個(gè)，三位數(shù)2^2過(guò)，……，k位數(shù)2^(k-1)個(gè)，
1+2+2^2+……+2^6=127,
2^k<=n<=2^k+2^(k-1)-1時(shí)I(n)=I[n-2^(k-1)]+1,
2^k+2^(k-1)<=n<=2^(k+1)-1時(shí)I(n)=I(n-2^k),
b1=2^0=1,
b2+b3=2^1+2^0=3,
b4+b5+b6+b7=2(b2+b3)+(b2+b3)=3(b2+b3)=9,
……
b64+b65+……+b127=3^6,
∴b1+b2+……+b127=1+3+3^2+……+3^6=(3^7-1)/2=1093.