c1 gãy tay

c2 cái bóng 

k mình nha

hay nhỉ mk k cho bạn đó

ko đc đăng các câu hỏi ko liên quan tới toán !

 \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)

\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2004}}\)

\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2005}}\)

\(=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2004}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2005}}\right)\)

\(=\frac{1}{2}-\frac{1}{2^{2005}}=\frac{2^{2004}-1}{2^{2005}}\)

\(A=\frac{2^{2004}-1}{2^{2005}}:\frac{1}{2}=\frac{2\left(2^{2004}-1\right)}{2^{2005}}=\frac{2^{2004}-1}{2^{2004}}\)

làm cách khác

đặt A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2004}}\)

2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2003}}\)

2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2003}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2004}}\right)\)

A = \(1-\frac{1}{2^{2004}}\)

đề bài sai bn ơi 

ih vuông góc vs ac 

vậy làm sao

dãy số trên ko có quy tắc:

\(\frac{1}{1}\)đến \(\frac{1}{7}\)= 1:7=\(\frac{1}{1}\):7

\(\frac{1}{7}\)đến \(\frac{1}{26}\)= ko có quy tắc 

vì vậy số hạn thứ 50 chưa thể tính