1. Biểu thức Scap S𝑆được viết lại thành S=3+(22−23+24−…−299+2100)cap S equals 3 plus open paren 2 squared minus 2 cubed plus 2 raised to the exponent 4 end-exponent minus … minus 2 raised to the exponent 99 end-exponent plus 2 raised to the exponent 100 end-exponent close paren𝑆=3+(22−23+24−…−299+2100).
2. Đặt A=22−23+24−…−299+2100cap A equals 2 squared minus 2 cubed plus 2 raised to the exponent 4 end-exponent minus … minus 2 raised to the exponent 99 end-exponent plus 2 raised to the exponent 100 end-exponent𝐴=22−23+24−…−299+2100.
3. Nhân Acap A𝐴với 222để có 2A=23−24+25−…−2100+21012 cap A equals 2 cubed minus 2 raised to the exponent 4 end-exponent plus 2 raised to the exponent 5 end-exponent minus … minus 2 raised to the exponent 100 end-exponent plus 2 raised to the exponent 101 end-exponent2𝐴=23−24+25−…−2100+2101.
4. Cộng Acap A𝐴và 2A2 cap A2𝐴để có 3A=(22−23+24−…−299+2100)+(23−24+25−…−2100+2101)3 cap A equals open paren 2 squared minus 2 cubed plus 2 raised to the exponent 4 end-exponent minus … minus 2 raised to the exponent 99 end-exponent plus 2 raised to the exponent 100 end-exponent close paren plus open paren 2 cubed minus 2 raised to the exponent 4 end-exponent plus 2 raised to the exponent 5 end-exponent minus … minus 2 raised to the exponent 100 end-exponent plus 2 raised to the exponent 101 end-exponent close paren3𝐴=(22−23+24−…−299+2100)+(23−24+25−…−2100+2101).
5. Rút gọn biểu thức 3A3 cap A3𝐴để có 3A=22+21013 cap A equals 2 squared plus 2 raised to the exponent 101 end-exponent3𝐴=22+2101.
6. Tính giá trị của Acap A𝐴bằng cách chia 3A3 cap A3𝐴cho 333.
7. Thay giá trị của Acap A𝐴vào biểu thức S=3+Acap S equals 3 plus cap A𝑆=3+𝐴để tìm tổng Scap S𝑆.

1. Biểu thức Acap A𝐴được viết là A=22−23+24−…−299+2100cap A equals 2 squared minus 2 cubed plus 2 raised to the exponent 4 end-exponent minus … minus 2 raised to the exponent 99 end-exponent plus 2 raised to the exponent 100 end-exponent𝐴=22−23+24−…−299+2100.
2. Nhân Acap A𝐴với 222ta có 2A=2(22−23+24−…−299+2100)=23−24+25−…−2100+21012 cap A equals 2 open paren 2 squared minus 2 cubed plus 2 raised to the exponent 4 end-exponent minus … minus 2 raised to the exponent 99 end-exponent plus 2 raised to the exponent 100 end-exponent close paren equals 2 cubed minus 2 raised to the exponent 4 end-exponent plus 2 raised to the exponent 5 end-exponent minus … minus 2 raised to the exponent 100 end-exponent plus 2 raised to the exponent 101 end-exponent2𝐴=2(22−23+24−…−299+2100)=23−24+25−…−2100+2101.
3. Cộng Acap A𝐴và 2A2 cap A2𝐴ta có A+2A=(22−23+24−…−299+2100)+(23−24+25−…−2100+2101)cap A plus 2 cap A equals open paren 2 squared minus 2 cubed plus 2 raised to the exponent 4 end-exponent minus … minus 2 raised to the exponent 99 end-exponent plus 2 raised to the exponent 100 end-exponent close paren plus open paren 2 cubed minus 2 raised to the exponent 4 end-exponent plus 2 raised to the exponent 5 end-exponent minus … minus 2 raised to the exponent 100 end-exponent plus 2 raised to the exponent 101 end-exponent close paren𝐴+2𝐴=(22−23+24−…−299+2100)+(23−24+25−…−2100+2101).
4. Rút gọn biểu thức ta có 3A=22+21013 cap A equals 2 squared plus 2 raised to the exponent 101 end-exponent3𝐴=22+2101.
5. Giá trị của Acap A𝐴được tính là A=22+21013=4+21013cap A equals the fraction with numerator 2 squared plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction equals the fraction with numerator 4 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝐴=22+21013=4+21013.
6. Thay giá trị của Acap A𝐴vào biểu thức S=3+Acap S equals 3 plus cap A𝑆=3+𝐴ta có S=3+4+21013cap S equals 3 plus the fraction with numerator 4 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝑆=3+4+21013.
7. Quy đồng mẫu số ta có S=3⋅3+4+21013=9+4+21013=13+21013cap S equals the fraction with numerator 3 center dot 3 plus 4 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction equals the fraction with numerator 9 plus 4 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction equals the fraction with numerator 13 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝑆=3⋅3+4+21013=9+4+21013=13+21013.

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lonh?

• Số lượng phần tử ban đầu là N=32cap N equals 32𝑁=32.
• Số lần so sánh tối đa được tính bằng công thức log2(N)log sub 2 open paren cap N close parenlog2(𝑁).
• Thay N=32cap N equals 32𝑁=32vào công thức, ta có log2(32)log sub 2 open paren 32 close parenlog2(32).
• Giá trị của log2(32)log sub 2 open paren 32 close parenlog2(32)là 555.
• Do đó, số lần so sánh tối đa là 555.

Lịch sử

no

12+14=26