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hơi khó đó tick mình nha Hoàng Thu Hà

Thuật toán: 

Bước 1: Nhập n

Bước 2: i←1; a←0;

Bước 3: a←a+1/(i*(i+2));

Bước 4: i←i+1;

Bước 5: Nếu i<=n thì quay lại bước 3

Bước 6: xuất a

Bước 7: Kết thúc

Viết chương trình:

uses crt;

var a:real;

i,n:longint;

begin

clrscr;

write('Nhap n='); readln(n);

a:=0;

for i:=1 to n do

a:=a+1/(i*(i+2));

writeln(a:4:2);

readln;

end.

Ta có:

\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)

\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)

\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)

...

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

=>

\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)

\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)

Vậy B < 2 

\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)

\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)

\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)

\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)

\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)

\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))

Vậy B < 2

2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)

2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)

2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)

2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)

2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)

2A = \(\frac{4032}{2017}\)

A = \(\frac{4032}{2017}\)\(:2\)

A = \(\frac{2016}{2017}\)

\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)

\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)

\(=\frac{1}{2}.2016.\frac{2}{2017}=\frac{2016}{2017}\)