Câu hỏi của Lê Phú Khánh

Question

Chung to rang 2¹+2²+2³+........+2¹⁰⁰ chia het cho 3

le bao truc · Accepted Answer

$S=2^1+2^2+2^3+...+2^{100}$
$S=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)$
$S=\left(2^1+2^2\right)+2^2\left(2^1+2^2\right)+...+2^{98}\left(2^1+2^2\right)$
$S=\left(2^1+2^2\right).\left(1+2^2+...+2^{98}\right)$
$S=6.\left(1+2^2+...+2^{98}\right)⋮3$

Câu trả lời:

$S=2^1+2^2+2^3+...+2^{100}$
$S=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)$
$S=\left(2^1+2^2\right)+2^2\left(2^1+2^2\right)+...+2^{98}\left(2^1+2^2\right)$
$S=\left(2^1+2^2\right).\left(1+2^2+...+2^{98}\right)$
$S=6.\left(1+2^2+...+2^{98}\right)⋮3$

Nguyen Viet Bac · Answer

S=21+22+23+...+2100$S=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)$S=(21+22)+(23+24)+...+(299+2100)$S=\left(2^1+2^2\right)+2^2\left(2^1+2^2\right)+...+2^{98}\left(2^1+2^2\right)$

Câu trả lời:

S=21+22+23+...+2100$S=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)$S=(21+22)+(23+24)+...+(299+2100)$S=\left(2^1+2^2\right)+2^2\left(2^1+2^2\right)+...+2^{98}\left(2^1+2^2\right)$

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