#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;

}

tham khảo:

\(a) 2+5+8+...+(3n−1)=n(3n+1)2 (1) Đặt Sn=2+5+8+...+(3n−1) Với n=1 ta có: S1=2=1(3.1+1)2 Giả sử (1) đúng với n=k(k≥1), tức là Sk=2+5+8+...+(3k−1)=k(3k+1)2 Ta chứng minh (1) đúng với n=k+1 hay Sk+1=(k+1)(3k+4)2 Thật vậy ta có: Sk+1=2+5+8+...+(3k−1)+[3(k+1)−1]=Sk+3k+2=k(3k+1)2+3k+2=3k2+k+6k+42=3k2+7k+42=(k+1)(3k+4)2 Vậy (1) đúng với mọi k≥1 hay (1) đúng với mọi n∈N∗ b) 3+9+27+...+3n=12(3n+1−3) (2) Đặt Sn=3+9+27+...+3n=12(3n+1−3) Với n=1, ta có: S1=3=12(32−3) (hệ thức đúng) Giả sử (2) đúng với n=k(k≥1) tức là Sk=3+9+27+...+3k=12(3k+1−3) Ta chứng minh (2) đúng với n=k+1, tức là chứng minh Sk+1=12(3k+2−3) Thật vậy, ta có: Sk+1=3+9+27+...+3k+1=Sk+3k+1=12(3k+1−3)+3k+1=32.3k+1−32=12(3k+2−3)(đpcm) Vậy (2) đúng với mọi k≥1 hay đúng với mọi n∈N∗\)

1:

\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^5\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^5}\right)}{n^3\left(1-\dfrac{2}{n^2}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}n^2\cdot3=+\infty\)

2: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{3n^6+3n^4-1}{3n-2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^6\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^6}\right)}{n\left(3-\dfrac{2}{n}\right)}=\lim\limits_{n\rightarrow\infty}n^5=+\infty\)

\(a,lim\dfrac{2n^2+1}{3n^3-3n+3}\)

\(=lim\dfrac{\dfrac{2}{n}+\dfrac{1}{n^3}}{3-\dfrac{3}{n^2}+\dfrac{3}{n^3}}=0\)

\(\lim\dfrac{-3n^3+1}{2n+5}=\lim\dfrac{-3n^2+\dfrac{1}{n}}{2+\dfrac{5}{n}}=\dfrac{-\infty}{2}=-\infty\)

\(\lim\dfrac{n^3-2n+1}{-3n-4}=\lim\dfrac{n^2-2+\dfrac{1}{n}}{-3-\dfrac{4}{n}}=\dfrac{+\infty}{-3}=-\infty\)

1: 

\(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^3+3n^2-1}{n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^3\left(-3+\dfrac{3}{n}-\dfrac{1}{n^3}\right)}{n^2\left(1-\dfrac{2}{n}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-3n^3}{n^2}=\lim\limits_{n\rightarrow\infty}-3n=-\infty\)

2: 

\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2-1}{-2n+3}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3-\dfrac{1}{n^2}\right)}{n\left(-2+\dfrac{3}{n}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-3}{2}n=-\infty\)

Bn viết để sai rồi, mk sửa lại :)

\(S\left(n\right)=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)

\(\Leftrightarrow3S\left(n\right)=\dfrac{1}{2}-\dfrac{1}{3n+2}\)

\(\Leftrightarrow S\left(n\right)=\dfrac{\dfrac{1}{2}-\dfrac{1}{3n+2}}{3}\)

I.

Do \(\left(u_n\right)\) là cấp số nhân \(\Rightarrow\)\(u_4=u_3.q\Rightarrow q=\dfrac{u_4}{u_3}=\dfrac{10}{3}\)

\(u_3=u_1q^2\Rightarrow u_1=\dfrac{u_3}{q^2}=\dfrac{27}{100}\)

2. Công thức số hạng tổng quát: \(u_n=\dfrac{27}{100}.\left(\dfrac{10}{3}\right)^{n-1}\)

II.

1. \(\lim\limits\dfrac{-3n^2+2n-2022}{3n^2-2022}=\lim\dfrac{-3+\dfrac{2}{n}-\dfrac{2022}{n^2}}{3-\dfrac{2022}{n^2}}=\dfrac{-3+0-0}{3-0}=-1\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2-5x+6}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x-3\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(x-3\right)=-1\)

\(b,lim\dfrac{2n^2+1}{3n^3-3n+3}\)

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

\(=n\times\dfrac{2}{3}=\)+∞

A, 7.b dương vô cực

\(a,lim\dfrac{7n^2-3n}{n^2+2}\)

\(=lim\dfrac{7-\dfrac{3}{n}}{1+\dfrac{2}{n^2}}\)

\(=\dfrac{7-0}{1+0}=\dfrac{7}{1}=7\)

`a)lim[5n^3-3n^2+1]/[1-3n^3]`

`=lim[5-3/n+1/[n^3]]/[1/[n^3]-3]`

`=5/[-3]=-5/3`

_____________________________
`b)lim[-9n+5]/[3n-3]`

`=lim[-9+5/n]/[3-3/n]`

`=[-9]/3=-3`

\(u_n=3n+1\left(n\in N^{\cdot}\right)\) là công thức tổng quát của dãy \(\left(u_n\right)\) mà mỗi số hạng của nó là số tự nhiên chia hết cho 3 dư 1 nên chọn câu A