1.
a, $S=1-3+3^2-3^3+...+3^{98}-3^{99}$
$⇒S=(1-3+3^2-3^3)+...+3^{96}(1-3+3^2-3^3)$
$⇒S=-20+3^{96}.(-20)$
$⇒-20(1+...+3^{96}$$\vdots$-20
b, $S=1-3+3^2-3^3+...+3^98-3^99$
$⇒-3S=(-3)+(-3)^2+...+(-3)^{100}$
$⇒S-(-3S)=1-(-3)^{100}$
$⇒4S=3^{100}+1$
$⇒S=$$\frac{3^{100}+1}{4}$
$⇒3^{100}:4$ dư 1
2:
a)4n-5$\vdots$n
⇒n∈Ư(5)={±1;±5}
b, Ta có: 3n+2$\vdots$2n-1
⇒2(3n+2)$\vdots$2n-1
⇒6n+4$\vdots$2n-1
⇒3(2n-1)+7$\vdots$2n-1
⇒2n-1∈Ư(7)={±1;±7}
2n-1 1 -1 7 -7
2n 2 0 8 -6
n 1 0 4 -3
Vậy n∈{1;0;4;-3}