\(1)B=1+3+3^2+...+3^{1999}+3^{2000}\\3B=3+3^2+3^3+...+3^{2000}+3^{2001}\\3B-B=3+3^2+3^3+...+3^{2000}+3^{2001}-(1+3+3^2+...+3^{1999}+3^{2000})\\2B=3^{2001}-1\\\Rightarrow B=\dfrac{3^{2001}-1}{2}\\---\)
\(2)C=1+4+4^2+...+4^{99}+4^{100}\\4C=4+4^2+4^3+...+4^{100}+4^{101}\\4C-C=4+4^2+4^3+...+4^{100}+4^{101}-(1+4+4^2+....+4^{99}+4^{100})\\3C=4^{101}-1\\\Rightarrow C=\dfrac{4^{101}-1}{3}\)
#\(Toru\)
1) \(B=1+3+3^2+...+3^{1999}+3^{2000}\)
\(3B=3\cdot\left(1+3+3^2+...+3^{2000}\right)\)
\(3B=3+3^2+...+3^{2001}\)
\(3B-B=3+3^2+3^3+...+3^{2001}-1-3-3^2-...-3^{2000}\)
\(2B=3^{2001}-1\)
\(B=\dfrac{3^{2001}-1}{2}\)
2) \(C=1+4+4^2+...+4^{100}\)
\(4C=4\cdot\left(1+4+4^2+...+4^{100}\right)\)
\(4C=4+4^2+4^3+...+4^{101}\)
\(4C-C=4+4^2+4^3+...+4^{201}-1-4-4^2-....-4^{100}\)
\(3C=4^{101}-1\)
\(C=\dfrac{4^{101}-1}{3}\)
Mình cho bạn công thức tổng quát để sau này tiện áp dụng nhé:
\(A=1+a^1+a^2+...+a^n\)
\(\Rightarrow A=\dfrac{a^{n+1}-1}{a-1}\)
