Question

Tính tổng S = \(1+5^2+5^4+...+5^{200}\)

Answer 1

Lời giải:

\(S=1+5^2+5^4+....+5^{198}+5^{200}\) (1)

\(\Rightarrow 5^2.S=5^2+5^4+...+5^{200}+5^{202}\) (2)

Lấy (2) trừ (1):

\(S(5^2-1)=(5^2+5^4+...+5^{200}+5^{202})-(1+5^2+....+5^{200})\)

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

Answer 2

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

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

\(5^2S=5^2+5^4+5^6+...+5^{202}.\)

\(5^2S-S=\left(5^2+5^4+5^6+...+5^{202}\right)-\left(1+5^2+5^4+...+5^{200}\right).\)

\(24S=5^{202}-1\Rightarrow S=\dfrac{5^{202}-1}{24}.\)

Vậy.....

Answer 3

Ta có:

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

<=> \(25S=5^2+5^4+5^6+...+5^{202}\)

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

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

Answer 4

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

25S=\(5^2+5^4+...+5^{202}\)

25S-S=\(\left(5^2+5^4+..+5^{202}\right)-\left(1+5^2+..+5^{200}\right)\)

\(\Rightarrow24S=5^{202}-1\)

\(\Rightarrow\)S=\(\dfrac{5^{202}-1}{24}\)

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