Question

A=1+4+4²+....+4¹⁰⁰

Answer 1

Lời giải:
\(A=1+4+4^2+4^3+...+4^{100}(1)\)

\(\Rightarrow 4A=4+4^2+4^3+...+4^{100}+4^{101}(2)\)

Lấy $(2)-(1)$ thu được:

\(3A=(4+4^2+4^3+...+4^{101})-(1+4+4^2+...+4^{100})\)

\(=4^{101}-1\)

\(\Rightarrow A=\frac{4^{101}-1}{3}\)

Answer 2

\(A=1+4+4^2+4^3+...+4^{100}\left(1\right)\)

=>\(4A=4+4^2+4^3+...+4^{100}+4^{101}\left(2\right)\)

lấy \(\left(2\right)-\left(1\right)\) ta được

Answer 3

bổ sung thêm:

\(3A=\left(4+4^2+4^3+...+4^{101}\right)-\left(1+4+4^2+...+4^{100}\right)\)

Answer 4

bổ sung thêm:

\(=4^{101}-1\)

=>\(A=\dfrac{4^{101}-1}{3}\)