Question

 Tính( rút gọn)

 C=1+1/2+1/2²+1/2³+...+1/2¹⁰⁰

 ^{tinh giup minh nha}

 

Answer 1

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

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

                         \(2C-C=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+2^{99}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

                         \(\Rightarrow C=2-\frac{1}{2^{100}}\)

                   Ủng hộ mk nha bn !!! ^_^

Answer 2

2C= 2+1+1/2+1/2²+.....+1/2⁹⁹

2C-C= ( 2+1+1/2+1/2²+.....+1/2⁹⁹ ) - (1+1/2+1/2²+1/2³+.....+1/2¹⁰⁰)

C= 2 - 1/2¹⁰⁰

C=2

Answer 3

C=1+1/2+1/2²+1/2³+...+1/2¹⁰⁰

=>C/2=1/2+1/2²+1/2³+1/2⁴+...+1/2¹⁰¹

=>C-C/a=(1+1/2+1/2²+1/2³+...+1/2¹⁰⁰)-(1/2+1/2²+1/2³+1/2⁴+...+1/2¹⁰¹)

C/2=1-1/2¹⁰¹

=>C=2*2¹⁰¹-1/2¹⁰¹

=2¹⁰¹-1/2¹⁰⁰

Vậy C=2¹⁰¹/2¹⁰⁰