Lời giải:
a.

$S=3^0+3^2+3^4+...+3^{2002}$

$3^2S=3^2+3^4+3^6+...+3^{2004}$

$3^2S-S=(3^2+3^4+3^6+...+3^{2004})-(3^0+3^2+3^4+...+3^{2002})$

$8S=3^{2004}-3^0=3^{2004}-1$

$S=\frac{3^{2004}-1}{8}$
b.

$S=(3^0+3^2+3^4)+(3^6+3^8+3^{10})+....+(3^{1998}+3^{2000}+3^{2002})$

$=(3^0+3^2+3^4)+3^6(3^0+3^2+3^4)+....+3^{1998}(3^0+3^2+3^4)$

$=(3^0+3^2+3^4)(1+3^6+...+3^{1998})$

$=91(1+3^6+...+3^{1998})=7.13(1+3^6+...+3^{1998})\vdots 7$

Ta có đpcm.

b: \(S=\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)

\(=91\cdot\left(1+...+3^{1998}\right)⋮7\)

b: \(S=3^0+3^2+3^4+...+3^{2002}\)

\(=\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)

\(=91\cdot\left(1+...+3^{1998}\right)⋮7\)

giup minh voi

tham khảo

https://olm.vn/hoi-dap/detail/49371559502.html

cái này khó

Ta thấy : các số hạng trong tổng S đều \(>\frac{7}{35}\) 

\(\Rightarrow S>\frac{7}{35}+\frac{7}{35}+\frac{7}{35}+\frac{7}{35}+\frac{7}{35}\)

\(\Rightarrow S>\frac{35}{35}\) 

\(\Rightarrow S>1\) ( đpcm )

Ta có: \(S=1+3^2+3^4+3^6+...+3^{98}\)

\(=\left(1+3^2\right)+\left(3^4+3^6\right)+...+\left(3^{96}+3^{98}\right)\)

\(=10+3^4\cdot10+...+3^{96}\cdot10\)

\(=10\left(1+3^4+...+3^{96}\right)⋮10\)(ĐPCM)

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

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

\(3B-B=3^{2017}-1\)

\(B=\dfrac{3^{2017}-1}{2}\)

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

\(3\cdot B=3+3^2+3^3+...+3^{2016}+3^{2017}\)

\(3B-B=3+3^2+3^3+...+3^{2016}+3^{2017}-\left(1+3^1+3^2+...+3^{2016}\right)\)

\(2B=3^{2017}-1\)

\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)

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

\(3\cdot B=3+3^2+3^3+...+3^{2017}\)

\(3B-B=3+3^2+3^3+...+3^{2017}-\left(1+3+3^2+...+3^{2016}\right)\)

\(2B=3^{2017}-1\)

\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)