bn chắc đề đúng chứ?chổ (1/2)^99 đó,2 cái liền hả?

đề lấy y hệt từ violympic 

\(\frac{1}{2}\)+\(\frac{1^2}{2^2}\)+\(\frac{1^3}{2^3}\)+...+\(\frac{1^{98}}{2^{98}}\)+\(\frac{1^{99}}{2^{99}}\)

=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{99}}\)

=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{98}}\)-\(\frac{1}{2^{99}}\)  Còn lại tự làm nhá kết quả cuối cùng là 2⁹⁹-1/2⁹⁹

ai trả lời đúng k

có cách làm nữa nha

1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)

   A= \(\frac{100}{2}\)

   A=50

2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)

     B= \(\frac{1}{99}\)

\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)

     \(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)

     \(=\frac{100}{2}\)

       \(=50\)

\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)

     \(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)

       \(=-\frac{1}{99}\)

21=45

đặt A=1/2+(1/2)^2+(1/2)^3+...+(1/2)^98+(1/2)^99+(1/2)^99

=>A=1/2+1²/2²+1³/2³+...+1⁹⁸/2⁹⁸+1⁹⁹/2⁹⁹+1⁹⁹/2⁹⁹

=>A=1/2+1/2²+1/2³+...+1/2⁹⁸+1/2⁹⁹+1/2⁹⁹

=>2A-1/2⁹⁹=1+1/2+1/2²+...+1/2⁹⁸

=>(2A-1/2⁹⁹)-(A-1/2⁹⁹)=(1+1/2+1/2²+...+1/2⁹⁸)-(1/2+1/2²+1/2³+...+1/2⁹⁸+1/2⁹⁹)

=>(2A-1/2⁹⁹)-(A-1/2⁹⁹)=1-1/2⁹⁹

=>A=1-1/2⁹⁹ +1/2⁹⁹=1

vậy A=1

chắc thế

cái phân số cuối sai thì phải 

bài này z đó, mk chắc chắn k sai đề

\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}...\frac{1-98^2}{98^2}.\frac{1-99^2}{99^2}\)

\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{98^2-1}{98^2}.\frac{99^2-1}{99^2}\)

= \(\frac{\left(2-1\right).\left(2+1\right)}{2^2}.\frac{\left(3-1\right).\left(3+1\right)}{3^2}.\frac{\left(4-1\right).\left(4+1\right)}{4^2}...\frac{\left(98-1\right)\left(98+1\right)}{98^2}.\frac{\left(99-1\right)\left(99+1\right)}{99^2}\)

\(=\frac{\left(2-1\right).\left(3-1\right).\left(4-1\right)...\left(99-1\right)}{2.3.4...98.99}.\frac{\left(2+1\right).\left(3+1\right).\left(4+1\right)...\left(99+1\right)}{2.3.4...98.99}\)

\(=\frac{1.2.3....98}{2.3.4...98.99}.\frac{3.4.5...100}{2.3.4...98.99}\)

\(=\frac{1}{99}.\frac{100}{2}\)

\(=\frac{50}{99}\)

Chúc bạn học tốt !!!

toi la Hai

dij me thg lol  duy xuyen

Ta có:

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

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

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

Áp dụng vào bài toán ta được

\(A=\frac{2.1+1}{\left[1\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)

\(=1-\frac{1}{100^2}=\frac{9999}{10000}\)

1+1=3 :)))