1) Tính C

\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)

\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)

\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)

\(=1-\frac{1}{n!}\)

3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)

\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)

Thanks !!! 

2.a) Vào question 126036

b) Vào question 68660

a, Chắc xét hàm số tổng quát!

Xét hàm số tổng quát:

\(\dfrac{1}{\left(k+1\right)\sqrt{k}}=\dfrac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\dfrac{1}{k\left(k+1\right)}\right)\)

\(=\sqrt{k}\left[\sqrt{\dfrac{1}{k}}^2-\sqrt{\dfrac{1}{k+1}}^2\right]\)

\(=\sqrt{k}\left(\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)

\(=\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)

Vì \(\dfrac{\sqrt{k}}{\sqrt{k+1}}< 1\Rightarrow1+\dfrac{\sqrt{k}}{\sqrt{k+1}}< 2\)

Do đó \(\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)< 2.\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)

\(\Rightarrow\dfrac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\) (1)

Áp dụng điểu (1) ta được:

\(\dfrac{1}{2}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\right)\)

\(\dfrac{1}{3\sqrt{2}}< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)\)

...................................

\(\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+....+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)

Với mọi giá trị của \(n>0\) ta luôn có: \(\sqrt{n+1}>0\)

Do đó \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\) (đpcm)

b)

program hotrotinhoc;

var s: real;

i,n: byte;

function t(x: byte): longint;

var j: byte;

t1: longint;

begin

t1:=1;

for j:=1 to x do

t1:=t1*j;

t1:=t;

end;

begin

readln(n);

s:=0;

for i:=1 to n do

s:=s+1/t(i);

write(s:1:2);

readln

end.

c) Đề em ghi sai rồi thế này với đúng :

\(T=1+\frac{2}{2^2}+\frac{3}{3^2}+\frac{4}{4^2}+...+\frac{n}{n^2}\)

program hotrotinhoc;

var t: real;

n,i: byte;

begin

readln(n);

t:=0;

for i:=1 to n do

t:=t+i/(i*i);

write(t:1:2);

readln

end.

a)

uses crt;

var N,S,i : integer;

begin clrscr;

S:=1;

for i:= 1 to N do S:=S*i;

writeln('N!=',S);

readln

end.

Các cái kia tương tự :))

d)

program hotrotinhoc;

var i,n: byte;

s: real;

function mu(x: byte): longint;

var j : byte;

k: longint;

begin

k:=1;

for j:=1 to x do

k:=k*x;

k:=mu;

end;

begin

readln(n);

s:=0;

for i:=1 to n do

s:=s+1/mu(i);

write(s:1:2);

readln

end.

e)

program hotrotinhoc;

var s: real;

i,n: byte;

begin

readln(n);

s:=0;

for i:=1 to n do

s:=s+i/(i+1);

write(s:1:2);

readln

end.