Question

Rút gọn biểu thức:

\(A=75\left(4^{1993}+4^{1992}+....+4^2+5\right)+25\)

Answer 1

Ta có \(A=75\left(4^{1993}+4^{1992}+....+4+1\right)+25\)

\(\Leftrightarrow A=25\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)+25\)

Vận dụng hằng đẳng thức

\(a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+...+b^{n-1}\right)\)

Ta có

\(\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)=4^{1994}-1\)

\(\Rightarrow A=25\left(4^{1994}-1\right)+25\)

\(\Leftrightarrow A=25\cdot4^{1994}\)

Vậy \(A=25\cdot4^{1994}\)