Question

Tính tổng : S = 1 + 9 + 9^2 + ......... + 9^2017

 

Answer 1

\(S=1+9+9^2+....+9^{2017}\)

\(9S=9.\left(1+9+9^2+...+9^{2017}\right)\)

\(9S=9+9^2+9^3+...+9^{2018}\)

\(8S=9S-S=\left(9+9^2+9^3+...+9^{2018}\right)-\left(1+9+9^2+....+9^{2017}\right)\)

\(8S=9^{2018}-1\)

\(S=\left(9^{2018}-1\right)\div8=\frac{9^{2018}-1}{8}\)

Vậy S = \(\frac{9^{2018}-1}{8}\)

Answer 2

S = 1 + 9 + 9\(^2\)+ . . . + 9\(^{2017}\)

9S = 9 + 9\(^2\)+ 9\(^3\)+ . . . + 9\(^{2018}\)

S = [ 9 + 9\(^2\)+ 9\(^3\)+ . . . + 9\(^{2018}\)] - [ 1 + 9 + 9\(^2\)+ . . . + 9\(^{2017}\)

S = [ 9 - 9 ] + [ 9\(^2\)- 9\(^2\) ] + [ 9\(^3\)- 9\(^3\)] + . . . + [ 9\(^{2017}\)- 9\(^{2017}\)] + [ 9\(^{2018}\)- 1 ]

S = 9\(^{2018}\)- 1

Answer 3

\(S=1+9+9^2+...+9^{2017}\)

\(\Rightarrow9S=9+9^2+9^3+...+9^{2018}\)

\(\Rightarrow9S-S=\left(9+9^2+...+9^{2018}\right)-\left(1+9+...+9^{2017}\right)\)

\(\Rightarrow8S=9^{2018}-1\)

\(\Rightarrow S=\frac{9^{2018}-1}{8}\)

Answer 4

nhầm nhầm, 9S - S = 8S chớ ko phải S