Câu hỏi của Trần Anh Dũng

Question

Tính các tổng sau bằng cách hợp líA=2+2^2+2^3+2^4+...+2^100B=1+3+3^2+3^3+...+3^2009C=1+5+5^2+5^3+...+5^1998D=4+4^2+4^3+...+4^n

Pham Ngoc Khương · Accepted Answer

A= 2+2^2+2^3+2^4+...+2^1002A=2^2+2^3+2^4+2^5+...+2^100+2^1012A-A=(2^2+2^3+2^4+2^5+..+2^100+2^101)-(2+2^2+2^3+2^4+...+2^100)A=2^101-2A=2^100B=1+3+3^2+3^3+...+3^20093B=3+3^2+3^3+3^4+...+3^2009+3^20103B+1=(1+3+3^2+3^3+3^4+...+3^2009)+3^20103B+1=B+3^20102B+1=3^20102B=3^2010-1B=(3^2010-1):2C=1+5+5^2+5^3+...+5^19985C=5+5^2+5^3+5^4+...+5^1998+5^19995C+1=(1+5+5^2+5^265^4+...+5^1998)+5^19995C+1=C+5^19994C+1=5^19994C=5^1999-1C=(5^1999-1):5D=4+4^2+4^3+...+4^n4D=4^2+4^3+4^4+...+4^n+4^(n+1)4D+4=(4+4^2+4^3+4^4+...+4^n)+4^(n+1)4D+4=D+4^(n+1)3D+4=4^(n+1)3D=4^(n+1)-4D=(4^(n+1)-4):3Câu trả lời: A= 2+2^2+2^3+2^4+...+2^1002A=2^2+2^3+2^4+2^5+...+2^100+2^1012A-A=(2^2+2^3+2^4+2^5+..+2^100+2^101)-(2+2^2+2^3+2^4+...+2^100)A=2^101-2A=2^100B=1+3+3^2+3^3+...+3^20093B=3+3^2+3^3+3^4+...+3^2009+3^20103B+1=(1+3+3^2+3^3+3^4+...+3^2009)+3^20103B+1=B+3^20102B+1=3^20102B=3^2010-1B=(3^2010-1):2C=1+5+5^2+5^3+...+5^19985C=5+5^2+5^3+5^4+...+5^1998+5^19995C+1=(1+5+5^2+5^265^4+...+5^1998)+5^19995C+1=C+5^19994C+1=5^19994C=5^1999-1C=(5^1999-1):5D=4+4^2+4^3+...+4^n4D=4^2+4^3+4^4+...+4^n+4^(n+1)4D+4=(4+4^2+4^3+4^4+...+4^n)+4^(n+1)4D+4=D+4^(n+1)3D+4=4^(n+1)3D=4^(n+1)-4D=(4^(n+1)-4):3

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