S = (3^0/2 + 1/2) + (3^1/2 + 1/2) + (3²/2 + 1/2) + (3³/2 + 1/2) +..+ 3^(n-1)/2 + 1/2 

S = n.(1/2) + (1/2)[3^0 + 3^1 + 3² +...+ 3^(n-1)] 

S = n/2 + (3^n - 1)/4 = (3^n + 2n - 1)/4 

mình lớp 5 mong bạn thông cảm và

\(S=1+2+5+14+....+\frac{3^{x-1}+1}{2}\)

\(=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+.....+\frac{3^{x-1}+1}{2}\)

\(=\frac{\left(3^0+1\right)+\left(3^1+1\right)+\left(3^2+1\right)+.....+\left(3^{x-1}+1\right)}{2}\)

\(=\frac{\left(1+3+3^2+.....+3^{x-1}\right)+x}{2}\)

Đặt \(A=1+3+3^2+....+3^{x-1}\)

\(3A-A=\left(3+3^2+....+3^x\right)-\left(1+3+....+3^{x-1}\right)\)

\(2A=3^x-1\Rightarrow A=\frac{3^x-1}{2}\)

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

  S = (3⁰/2 + 1/2) + (3¹/2 + 1/2) + (3²/2 + 1/2) + (3³/2 + 1/2) +..+ 3^n-1/2 + 1/2 

S = n.(1/2) + (1/2)[3^0 + 3^1 + 3² +...+ 3^n-1] 

S = n/2 + (3^n - 1)/4 = (3^n + 2n - 1)/4 

S = (3⁰/2 + 1/2) + (3¹/2 + 1/2) + (3²/2 + 1/2) + (3³/2 + 1/2) +..+ 3^(n-1)/2 + 1/2 

S = n.(1/2) + (1/2)[3⁰ + 3¹ + 3² +...+ 3^(n-1)] 

S = n/2 + (3ⁿ - 1)/4 = (3ⁿ + 2n - 1)/4

A=1+2+5+…+(3^(n-1)+1)/2

\(S=\frac{3^{1-1}+1}{2}+\frac{3^{2-1}+1}{2}+\frac{3^{3-1}+1}{2}+...+\frac{3^{n-1}+1}{2}\)

\(S=\frac{\left(3^0+3^1+3^2+...+3^{n-1}\right)+\left(1+1+1+...+1\right)}{2}\)có n c/s 1

\(S=\frac{\frac{\left(3^n-1\right)}{2}+n}{2}\)

\(=3^n-1+\frac{n}{2}\)

...\(3^0+3^1+3^3+...+3^{n-1}\)bạn tính nha

Câu hỏi của WINNER - Toán lớp 7 - Học toán với OnlineMath

๖ۣۜƝƘ☆ŤŔầŃ➻❥VăŃ➻❥ŃÁM❖︵copy ở đây

\(S=1+2+5+14+...+\frac{3^{x-1}+1}{2}\)

\(\Rightarrow2S=2+4+10+28+...+\left(3^{x-1}+1\right)\)

\(\Rightarrow2S=\left[1+1+1+1+...+1\right]+\left[1+3+9+27+...+3^{x-1}\right]\)

Đặt : \(S1=1+1+1+1+...+1=x\)

\(S2=1+3+9+27+...+3^{x-1}\)

\(\Rightarrow3.S2=3+9+27+81+...+3^x\)

\(\Rightarrow3.S2=S2=2.S2=3^x-1\Leftrightarrow S2=\frac{3^x-1}{2}\)

\(\Rightarrow S=\frac{S1+S2}{2}=\frac{x+\frac{3^x-1}{2}}{2}\)

\(\Rightarrow S=\frac{3^x+2x-1}{4}\)

Vậy : \(S=\frac{3^x+2x-1}{4}\)

#include <bits/stdc++.h>

using namespace std;

int main()

{

int n,i,t=0;

cin>>n;

for (int i=1; i<=3*n+1; i++)

if (i%2==1) t+=i;

else t-=i;

cout<<t;

}

Câu 13

S = 1 + 2 + 2² + ... + 2¹⁰

2S = 2 + 2² + 2³ + ... + 2¹¹

S = 2S - S

= (2 + 2² + 2³ + ... + 2¹¹) - (1 + 2 + 2² + ... + 2¹⁰)

= 2¹¹ - 1

= 2048 - 1

= 2047

Câu 14

3n + 2 = 3n - 6 + 8 = 3(n - 2) + 8

Để (3n + 2) ⋮ (n - 2) thì 8 ⋮ (n - 2)

⇒ n - 2 ∈ Ư(8) = {-8; -4; -2; -1; 1; 2; 4; 8}

⇒ n ∈ {-6; -2; 0; 1; 3; 4; 6; 10}

Mà n là số tự nhiên

⇒ n ∈ {0; 1; 3; 4; 6; 10}