Question

Cho biểu thức:B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.....+\frac{3^{98}+1}{3^{98}}\).Chứng minh rằng B<100

Answer 1

B = \(\frac{4}{3^1}+\frac{10}{3^2}+\frac{28}{3^3}+...+\frac{3^{98}+1}{3^{98}}\)

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

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

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

=> B < 98 < 100

vậy B < 100

Answer 2

dễ thui chờ tí nhé

Answer 3

\(B=\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\)

\(3B=3\left(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\right)\)

\(3B=4+\frac{10}{3}+...+\frac{3^{98}+1}{3^{97}}\)

\(3B=\left(4+\frac{10}{3}+...+\frac{3^{98}+1}{3^{97}}\right)-\left(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\right)\)

\(2B=4-\frac{3^{98}+1}{3^{98}}\)

\(B=2-\frac{3^{98}+1}{2.3^{98}}<2\)

mà 2<100

=>B<100

Answer 4

chết tui sai forever Mathemas ms đúng

Answer 5

Deu giai  sai