Ta có : S₁ = 1 + (-3) + 5 + (-7) + .... + 17

                = (1 - 3) + (5 - 7) + (9 - 11)+ (13 - 15) + 17 

                = -2 + -2 + -2 + -2 + 17 

                = -2 x 4 + 17 

                = -8 + 17 

             S₁ = 9

S₂ = (4 - 2) + (8 - 6) + (12 - 10) + (16 - 14) + -18

     = 2 x 4 - 18 

S₂ = -10

S₁ + S₂ = 9 - 10 = -1

S₁=1+(-3)+5+(-7)+...+17.

S₁=-2+(-2)+....+(-2).(9 số -2).

S₂=-2+4+(-6)+....+(-18)

S₂=-2+(-2)+...+(-2).(9 số -2).

=> (-2).(9+9)=-36.

S₁ = 1 + (-3) + 5 + (-7) + ............+ 17

= [ 1+(-3) ] + [5+(-7) + .........+ 17

= -2   +  ( -2 ) + ......+ 17  ( có 4 số -2)

= -2  . 4  + 17

=  -8  + 17 = 9

S₂ = -2  + 4 + (-6) +8 ....+(-18)

= [ (-2)+4]  + [ (-6) + 8 + ......+ (-18) (có 4 cặp số )

= 2  + 2 + .....+ (-18)

= 2 . 4 + (-18)

= 8 + (-18)

= -10

suy ra S₁ + S₂  = 9 + (-10) = -1

#include<bits/stdc++.h>

using namespace std;

     int main() {

     int N;

     cin >> N;

     int sum = 0;

     for (int i = 1; i <= N; i++) {

          if (sqrt(i) == (int)sqrt(i)) {

               sum += i;

          }

     }

     cout << sum << endl;

     return 0;

}

\(S=1+3+3^2+3^3+3^4+.....+3^{16}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}\right)\)

\(=1\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+......+3^{2014}\left(1+3+3^2\right)\)

\(=1.13+3^3.13+.....+3^{2014}.13\)

\(=13\left(1+3^3+....+3^{2014}\right)⋮13\)

\(\Rightarrow S⋮13\)

\(S=7+7^2+7^3+...7^{20}\)

Ta có: \(7S=7.\left(7+7^2+7^3+...+7^{20}\right)\)

\(7S=7^2+7^3+7^4+...+7^{21}\)

\(7S-S=\left(7^2+7^3+7^4+...+7^{21}\right)-\left(7+7^2+7^3+...+7^{20}\right)\)

\(6S=\left(7^{21}-7\right)\)

\(S=\left(7^{21}-7\right):6\)

Chúc bạn học tốt

7S=7^2+7^3+...+7^21

=>6S=7^21-7

=>S=(7^21-7)/6

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

#include <bits/stdc++.h>

int main() {
    int n;
    cin>>n;
    int sum=0;

    for(int i=1;i<=n;i++)
    {
        sum+=i*i;
    }
    cout<<"The total is: "<<sum<<endl;

    for(int j=0;j<=50000;j++)
    {
        int du=j%10;
        int tongcacso=j%10*j%10*j%10;
        cout<<"du="<<du<<endl;
        sum=sum+du*du*du;
        cout<<"\nsum= "<<sum<<endl;
        cout<<"sum= (sum+j*j*j) "<<endl;
    }

    return 0;
}

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)

câu 1: số đó là :87

?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)

3A-A= \(1-\frac{1}{3^{2008}}\)

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

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

3B - B = \(1-\frac{1}{3^n}\)

Ta có :

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)

\(\Leftrightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow\)\(2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow\)\(2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{3^{2008}}:2\)

\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{2.3^{2008}}\)

Vậy \(A=\frac{3^{2008}-1}{2.3^{2008}}\)