M = 1 + 3 + 3^2 + ... + 3^25

3M = 3 + 3^2 + ... + 3^26

3M - M = ( 3 + 3^2 + ... + 3^26 ) - ( 1 + 3 + ... + 3^25 )

2M = 3^26 - 1 

M = ( 3^26 - 1 ) : 2

N - M = [( 3^26 - 1 ) : 2 ] - ( 3^26 : 2 )

N - M = 3^26 : 2 - 1 : 2 - 3^26 : 2

N - M = -1/2

a/ \(\lim\limits\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}=\lim\limits\dfrac{\dfrac{\left(\dfrac{1}{3}\right)^{n+1}-1}{\dfrac{1}{3}-1}}{\dfrac{\left(\dfrac{1}{2}\right)^{n+1}-1}{\dfrac{1}{2}-1}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}=3\)

b/ \(\lim\limits\left(n^3+n\sqrt{n}-5\right)=+\infty-5=+\infty\)

\(y=\dfrac{mx+2}{x+n}\left(x\ne-n\right)\)

Để hàm số có tiệm cận đứng x=2, thì mẫu có nghiệm x=2

\(\Leftrightarrow2+n=0\Leftrightarrow n=-2\)

\(A\left(3;-1\right)\in y\Rightarrow-1=\dfrac{3m+2}{3-2}\Rightarrow m=-1\)

\(\Rightarrow m+n=-1-2=-3\)

bn ơi đề kiu j vậy
 

\(\frac{1}{\sqrt{a-1}+\sqrt{a}}=\frac{\sqrt{a}-\sqrt{a-1}}{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}=\frac{\sqrt{a}-\sqrt{a-1}}{a-a+1}=\sqrt{a}-\sqrt{a-1}\)
Áp dụng cái này vào là ra bạn nhé

nhanh mn ơi

uses crt;

var n,i:integer;

s:real;

//chuongtrinhcon

function gthua(n:integer):real;

var gt:real;

i:integer;

begin 

gt:=1; 

for i:=1 to n do    

gt:=gt*i; 

 gthua:=gt;

end;

//chuongtrinhcon

function lthua(n,x:integer):real;

var lt:real;

i:integer;

begin

lt:=1;

for i:=1 to x do lt:=lt*x;

lthua:=lt;

end;

//chuongtrinhchinh

begin

clrscr;

readln(n);

s:=0;

for i:=1 to n do

s:=s+(lthua(i,i)/gthua(i));

writeln(s:4:2);

readln;

end.