Question

Tính giá trị của biểu thức:

A=(1/3+1,3^2+1/3^3+1/3^4).3^5+(1/3^5+1/3^6+1/3^7+1/3^8).3^9+...+(1/3^97+1/3^98+1/3^99+1/3^100).3^101

Answer 1

\(A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\cdot3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}\right)\cdot3^9+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\cdot3^{101}\)=\(\left(\frac{3^5}{3}+\frac{3^5}{3^2}+\frac{3^5}{3^3}+\frac{3^5}{3^4}\right)+\left(\frac{3^9}{3^5}+\frac{3^9}{3^6}+\frac{3^9}{3^7}+\frac{3^9}{3^8}\right)+...+\left(\frac{3^{101}}{3^{97}}+\frac{3^{101}}{3^{98}}+\frac{3^{101}}{3^{99}}+\frac{3^{101}}{3^{100}}\right)\)

=(3+3²+3³+3⁴)+(3+3²+3³+3⁴)+...+(3+3²+3³+3⁴)

Tổng trên có số số hạng là(mỗi ngoặc là 1 số hạng)

(101-5):4+1=25(số hạng)

=>A=25.(3+3²+3³+3⁴)=25.120=3000