Step 1: Đặt biểu thức Đặt biểu thức đã cho là Scap S𝑆: S=3+22−23+24−…−299+2100cap S equals 3 plus 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-exponent𝑆=3+22−23+24−…−299+2100 Step 2: Tách phần tổng Tách số 3 ra khỏi tổng và đặt phần còn lại là Acap A𝐴: 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 Khi đó S=3+Acap S equals 3 plus cap A𝑆=3+𝐴. Step 3: Nhân A với 2 Nhân biểu thức Acap A𝐴với 2: 2A=2(22−23+24−…−299+2100)2 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 paren2𝐴=2(22−23+24−…−299+2100) 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 Step 4: Cộng A và 2A Cộng Acap A𝐴và 2A2 cap A2𝐴lại với nhau. Các số hạng từ 232 cubed23đến 21002 raised to the exponent 100 end-exponent2100sẽ triệt tiêu: 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) 3A=22+21013 cap A equals 2 squared plus 2 raised to the exponent 101 end-exponent3𝐴=22+2101 3A=4+21013 cap A equals 4 plus 2 raised to the exponent 101 end-exponent3𝐴=4+2101 Step 5: Tìm A Tìm giá trị của Acap A𝐴: A=4+21013cap A equals the fraction with numerator 4 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝐴=4+21013 Step 6: Tìm S Thay Acap A𝐴vào biểu thức của Scap S𝑆: S=3+Acap S equals 3 plus cap A𝑆=3+𝐴 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 S=9+4+21013cap S equals the fraction with numerator 9 plus 4 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝑆=9+4+21013 S=13+21013cap S equals the fraction with numerator 13 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝑆=13+21013 Answer: S=13+21013bold cap S equals the fraction with numerator 13 plus 2 raised to the exponent 101 end-exponent and denominator 3 end-fraction𝐒=𝟏𝟑+𝟐𝟏𝟎𝟏𝟑 đấy coppy bạn bên dưới =)))

kbt nx =))))

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