1.Tính tổng:
S=1+9+\(9^2\)+...+\(9^{2017}\)
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S=2+4+6+...+98+100
S=\(\frac{\left[\left(\frac{100-2}{2}+1\right).\left(100+2\right)\right]}{2}=2550\)
S=1+2+3+4+...+2016+2017
S=\(\frac{\left(2017-1+1\right).\left(2017+1\right)}{2}=2035153\)
1.Số lượng số của S= (2017-1)+1=2017 số
tổng=(2016+1).(2016:2)+2017=2 035 153
2.Số lượng số của S=(100-2):2+1=50 số
tổng=(100+2).(50:2)=2 550
\(S=1+9+9^2+...+9^{2017}\)
\(\Rightarrow9S=9+9^2+9^3+...+9^{2018}\)
\(\Rightarrow9S-S=8S=\left(9+9^2+9^3+...+9^{2018}\right)-\left(1+9+9^2+...+9^{2017}\right)\)
\(8S=9^{2018}-1\)
\(\Rightarrow S=\frac{9^{2018}-1}{8}\)
\(S=3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\)
\(S\cdot\frac{1}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(S\cdot\frac{2}{3}=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(S\cdot\frac{2}{3}-S\cdot\frac{1}{3}=2+1+\frac{1}{2}+...+\frac{1}{2^8}-1-\frac{1}{2}-...-\frac{1}{2^9}\)
\(S\cdot\frac{1}{3}=2-\frac{1}{2^9}\)
\(S=\left(2-\frac{1}{2^9}\right):\frac{1}{3}\)
\(S=\left(2-\frac{1}{2^9}\right)\cdot3\)
\(S=6-\frac{3}{2^9}\)
\(S=\frac{6\cdot2^9-3}{2^9}\)
\(S=1+9+9^2+....+9^{2017}\)
\(9S=9.\left(1+9+9^2+...+9^{2017}\right)\)
\(9S=9+9^2+9^3+...+9^{2018}\)
\(8S=9S-S=\left(9+9^2+9^3+...+9^{2018}\right)-\left(1+9+9^2+....+9^{2017}\right)\)
\(8S=9^{2018}-1\)
\(S=\left(9^{2018}-1\right)\div8=\frac{9^{2018}-1}{8}\)
Vậy S = \(\frac{9^{2018}-1}{8}\)
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
S=1+9+92+93+...+92017
9S =(90+9+92+93+...+92017).2
9S =9+92+93+94+...+92017+92018
9S-S =(9+92+93+94+...+92017+92018)-(90+9+92+93+...+92017)
8S =92018-90
S =(92018-90):8
=> 9S=9+9^2+9^3+...+9^2018
=> 9S-S=8S=(9+9^2+9^3+...+9^2018)-(1+9+9^2+9^3+...+9^2017)
=> 8S=9+9^2+...+9^2018-1-9-9^2-...-9^2017
=> 8S=9^2018-1
=> S=(9^2018-1)/8