#include <conio.h>

#include <stdio.h>

int tong(int n)

{

int i;

int s=0;

for (i=0;i<=n;i++)

s+=(2*i+1);

return s;

}

void main()

{

int n;

printf("\nNhap N=   ");

scanf("%d",&n);

printf("\n Tong s = %d",tong(n));

getch();

}

a,S=1+3+3²+...+3⁶⁰

3S=3+3²+3³+...+3⁶¹

3S-S=(3+3²+3³+...+3⁶¹)-(1+3+3²+...+3⁶⁰)

2S = 3⁶¹ - 1

b,2S+1=3⁶¹-1+1=3⁶¹ = 3^x-3

=>x-3=61=>x=64

c, S=1+3+3²+...+3⁶⁰

=(1+3)+(3²+3³)+...+(3⁵⁹+3⁶⁰)

=4+3²(1+3)+...+3⁵⁹(1+3)

=4+3².4+...+3⁵⁹.4

=4(1+3²+...+3⁵⁹) chia hết cho 4

S=1+3+3²+...+3⁶⁰

=(1+3+3²)+....+(3⁵⁸+3⁵⁹+3⁶⁰)

=13+...+3⁵⁸(1+3+3²)

=13+...+3⁵⁸.13

=13(1+...+3⁵⁸)

còn S chia hết cho 10

S chia hết cho 40 mới c/m đc

a) Ta có: \(S=1+3+3^2+3^3+...+3^{98}\)

\(3S=3+3^2+3^3+3^4+...+3^{99}\)

\(3S-S=3^{99}-1\)

Hay \(2S=3^{99}-1\)

\(\Rightarrow S=\frac{3^{99}-1}{2}\)

b) Ta có: \(2S=3^{5x-1}-1\)

\(\Rightarrow3^{99}-1=3^{5x-1}-1\)

\(\Rightarrow3^{99}=3^{5x-1}\)

\(\Rightarrow5x-1=99\)

\(\Rightarrow5x=100\)

\(\Rightarrow x=20\)

Hok tốt nha^^

Câu 1: A

Câu 2: C

\(A,\)\(S=\left(3+3^2\right)+\left(3+3^2\right)3^2+...+\left(3+3^2\right)3^{2018} \)

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

\(\Rightarrow S⋮9\)

\(B,\)\(S=3+3^2+3^3+\left(3+3^2+3^3\right)3^3+...\left(3+3^2+3^3\right)3^{2017}\)

\(S=39+39.3^3+...+39.3^{2017}\)

Nhưng xét lại thì thấy 2017 không chia hết cho 3 nên câu b có lẽ sai đề =)))))

\(C,\)\(S=\left(1+3+3^2+3^3\right).3+\left(1+3+3^2+3^3\right).3^4+...+\left(1+3+3^2+3^3\right).3^{2017}\)

\(S=40.3+40.3^4+...+40.3^{2017}\)

\(Vậy...\)

a/

\(3S=3+3^2+3^3+3^4+...+3^{120}\)

\(2S=3S-S=3^{120}-1\Rightarrow S=\frac{3^{120}-1}{2}\)

b/ \(S=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)

\(S=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

\(S=13+3^3.13+...+3^{117}.13=13\left(1+3^3+...+3^{117}\right)\) chia hết cho 13

c/

\(S=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)

\(S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)

\(S=40+3^4.40+...+3^{116}.40=40\left(1+3^4+...+3^{116}\right)\) chia hết cho 40

1. C

2. D

3. C

4. D

5. D

6. B

7. D

8. B

9. C

Lời giải:

\(a+b=3\Rightarrow a+(b-2)=1\Rightarrow b-2=1-a\)

Ta có:

\(f(x)=\frac{9^x}{9^x+3}\Rightarrow f(a)=\frac{9^a}{9^a+3}\) (1)

\(f(b-2)=f(1-a)=\frac{9^{1-a}}{9^{1-a}+3}=\frac{9}{9^a\left(\frac{9}{9^a}+3\right)}\)

\(=\frac{9}{9+3.9^a}=\frac{3}{3+9^a}\) (2)

Từ (1),(2) suy ra \(f(a)+f(b-2)=\frac{9^a}{9^a+3}+\frac{3}{3+9^a}=\frac{9^a+3}{9^a+3}=1\)

Đáp án A