n bằng 100

ta có: \(\frac{31+32+35}{34}=\frac{31}{34}+\frac{32}{34}+\frac{35}{34}.\)

mà \(\frac{31}{32}>\frac{31}{34};\frac{32}{33}>\frac{32}{34}\)

\(\Rightarrow\frac{31}{32}+\frac{32}{33}+\frac{35}{34}>\frac{31}{34}+\frac{32}{34}+\frac{35}{34}=\frac{31+32+35}{34}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(\Leftrightarrow S=1\left(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S-S=1+\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(\Leftrightarrow S=1-\frac{1}{60}=\frac{59}{60}\)

(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)(1/51 +1/52+  1/53 +1/54+ 1/55 +1/56+  1/57 +1/58 + 1/59 +1/60)

E = 1/31+1/32+...+1/60

E > 1/40+1/40+...+1/40+1/41+1/42+...+1/60

E > 20/40+1/41+1/42+...+1/60

E > 1/2+1/60+1/60+...+1/60

E > 1/2 + 1/3 = 5/6

Mà 5/6 > 4/5

=> E > 4/5

Lời giải:

$A=1+3+3^2+3^3+....+3^{30}$

$3A=3+3^2+3^3+....+3^{31}$

$3A-A=(3+3^2+3^3+...+3^{31})-(1+3+...+3^{30})$

$2A=3^{31}-1$

$A=\frac{3^{31}-1}{2}=\frac{3.3^{30}-1}{2}$

$=\frac{3.9^{15}-1}{2}$

Ta thấy: Đối với $9^n$ thì $n$ chẵn số này sẽ có tận cùng là $1$, $n$ lẻ sẽ có tận cùng là $9$

Vậy $9^{15}$ tận cùng là $9$

$\Rightarrow 3.9^{15}$ tận cùng là $7$

$\Rightarrow 3.9^{15}-1$ tận cùng là $6$

$\Rightarrow A=\frac{3.9^{15}-1}{2}$ tận cùng là $3$ hoặc $8$

Do đó $A$ không thể là scp.

\(B=3^1+3^2+3^3+...+3^{300}\\=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{299}+3^{300})\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+...+3^{299}\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{299}\cdot4\\=4\cdot(3+3^3+3^5+...+3^{299})\)

Vì \(4\cdot(3+3^3+3^5+...+3^{299})\vdots2\)

nên \(B\vdots2\)

B=(3+3²)+(3³+3⁴)+...+(3²⁹⁹+3³⁰⁰)

B=3(1+3)+3³(1+3)+...+3²⁹⁹(1+3)

B=3.4+3³.4+...+3^299.4

B=4(3+3³+...+3²⁹⁹) chia hết cho 2 vì 4 chia hết cho 2

vậy B chia hết cho 2

Camr own mn