Ta có : A= x^0+ x^1+ x^2+...+x^n => \(A=\frac{x^{n+1}-1}{x-1}\)

Chứng minh: xA=x¹+x²+...+x^ⁿ⁺¹

xA-A=A(x-1)=xⁿ⁺¹-x⁰=xⁿ⁺¹-1

Từ đó => điều trên

Vậy Ta có:

\(S=\frac{\left(-\frac{1}{7}\right)^{2017}-1}{-\frac{1}{7}-1}\)

A = (1-2-3+4)+(5-6-7+8)+.....+(2013-2014-2015+2016)

   = 0+0+.....+0

   = 0

Tk mk nha

Câu 1 :

 \(A=5+5^2+5^3+......+5^{2016}\)( 1 )

\(A.5=5^2+5^3+5^4+.....+5^{2017}\)( 2 )

Lấy ( 2 ) - ( 1 ) ta được :

\(A.5-A=\left(5^2+5^3+5^4+......+5^{2017}\right)-\left(5+5^2+5^3+......+5^{2016}\right)\)

          \(A.4=5^{2017}-5\)

              \(A=\left(5^{2017}-5\right):4\)

\(\Rightarrow4A+5=\left(5^{2017}-5\right):4.4\)

\(\Rightarrow4A+5=5^{2017}-5\)

Câu 2 :

\(A=7+7^2+7^3+.....+7^{2016}\)( 1 )

\(A.7=7^2+7^3+7^4+......+7^{2017}\)( 2 )

Lấy ( 2 ) - ( 1 ) ta được :

\(A.7-A=\left(7^2+7^3+7^4+.....+7^{2017}\right)-\left(7+7^2+7^3+.....+7^{2016}\right)\)

          \(A.6=7^{2017}-7\)

              \(A=\left(7^{2017}-7\right):6\)

\(\Rightarrow6.A+7=\left(7^{2017}-7\right):6.6\)

\(\Rightarrow6.A+7=7^{2017}-7\)

Câu 1:

\(A=5+5^2+5^3+......+5^{2016}\)

\(5A=5^2+5^3+5^4+.......+5^{2017}\)

\(\Rightarrow5A-A=\left(5^2+5^3+5^4+.......+5^{2017}\right)-\left(5+5^2+5^3+.......+5^{2016}\right)\)

\(\Rightarrow5A-A=5^2+5^3+5^4+.........+5^{2017}-5-5^2-5^3-........-5^{2016}\)

\(\Rightarrow4A=5^{2017}-5\)

\(\Rightarrow4A+5=5^{2017}\)

Câu 2 tương tự câu 1

Bạn đánh sai đề à. Sao ko có quy luật gì vậy?