`G=1-5+5^{2}-5^{3}+5^{4}-...-5^{199}-5^{200}`

`5G=5-5^{2}+5^{3}-5^{4}+5^{5}-...-5^{200}-5^{201}`

`=>5G+G=1-2.5^{200}-5^{201}`

\(=>G=\dfrac{1-2.5^{200}-5^{201}}{6}\)

Lời giải:
Gọi tổng trên là $K$
$K=1+5^2+5^3+5^4+...+5^{200}$

$5K=5+5^3+5^4+5^5+...+5^{201}$
$\Rightarrow 5K-K = 5+5^{201}-1-5^2$

$\Rightarrow 4K = 5^{201}-21$

$\Rightarrow K= \frac{5^{201}-21}{4}$

Hồ Đại Tiến ngu đề là gì

S=1+5²+5⁴+…+5²⁰⁰

=>5².S=5²+5⁴+5⁶+…+5²⁰²

=>25.S-S=5²+5⁴+5⁶+…+5²⁰²-1-5²-5⁴-…-5²⁰⁰

=>24.S=5²⁰²-1

=>\(S=\frac{5^{202}-1}{24}\)

chán!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(C=1+5^2+5^3+5^4+...+5^{200}\)

\(\Rightarrow C+5=1+5+5^2+5^3+5^4+...+5^{200}\)

\(\Rightarrow C+5=\dfrac{5^{200+1}-1}{5-1}\)

\(\Rightarrow C+5=\dfrac{5^{201}-1}{4}\)

\(\Rightarrow C=\dfrac{5^{201}-1}{4}-5\)

\(\Rightarrow C=\dfrac{5^{201}-21}{4}\)

\(S=1+5+5^2+5^3+...+5^{200}\)

\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{201}\)

\(\Rightarrow5S-S=\left(5+5^2+5^3+...+5^{201}\right)-\left(1+5+5^2+...+5^{200}\right)\)

\(\Rightarrow4S=5^{201}-1\)

\(\Rightarrow S=\frac{5^{201}-1}{4}\)

\(S=1+5^2+...+5^{200}\)

\(5S=5+5^3+...+5^{201}\)

\(5S-S=\left(5+5^3+...+5^{201}\right)-\left(1+5^2+...+5^{200}\right)\)

\(4S=5+5^{201}-1+5^2\)

\(4S=5^{201}+29\)

\(S=\frac{5^{201}+29}{4}\)

\(S=1+5^2+5^4+...+5^{200}\)
\(25S=5^2+5^4+5^6+...+5^{202}\)
\(\Rightarrow25S-S=5^{202}-1\)
\(\Leftrightarrow S=\frac{5^{202}-1}{24}\)