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Bài 1:
\(A=7+7^3+7^5+...+7^{1999}\)
\(\Rightarrow A=\left(7+7^3\right)+\left(7^5+7^7\right)+...+\left(7^{1997}+7^{1999}\right)\)
\(\Rightarrow A=\left(7+343\right)+7^4\left(7+7^3\right)+...+7^{1996}\left(7+7^3\right)\)
\(\Rightarrow A=350+7^4.350+...+7^{1996}.350\)
\(\Rightarrow A=\left(1+7^4+...+7^{1996}\right).350⋮35\)
\(\Rightarrow A⋮35\left(đpcm\right)\)
b2:
a) \(S=1+3+3^2+...+3^{49}\)
\(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{48}+3^{49}\right)\)
\(\Rightarrow S=\left(1+3\right)+3^2\left(1+3\right)+...+3^{48}\left(1+3\right)\)
\(\Rightarrow S=4+3^2.4+...+3^{48}.4\)
\(\Rightarrow S=\left(1+3^2+...+3^{48}\right).4⋮4\)
\(\Rightarrow S⋮4\left(đpcm\right)\)
c) \(S=1+3+3^2+...+3^{49}\)
\(\Rightarrow3S=3+3^2+3^3+...+3^{50}\)
\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{50}\right)-\left(1+3+3^2+...+3^{49}\right)\)
\(\Rightarrow2S=3^{50}-1\)
\(\Rightarrow S=\frac{3^{50}-1}{2}\left(đpcm\right)\)
D = 2 . 3 + 4 . 5 + 6. 7 + ... + 50 . 51 = 150,850
E = 1 . 99 + 2 . 98 + 3 . 97 + ... + 49 . 51 + 50 .50 = 82,464
Tick nha
đặt A=................................
\(7A=7+7^2+7^3+.........+7^{51}.\)
\(7A-A=7^{51}-1\)
\(A=\frac{7^{51}-1}{6}\)
chúc bn vui vẻ thành công trong năm mới 2018
\(M=\left(\dfrac{1}{50}+\dfrac{2}{49}+...+\dfrac{50}{1}\right):\left(\dfrac{7}{2}+\dfrac{7}{3}+...+\dfrac{7}{50}+\dfrac{7}{51}\right)\)
\(=\dfrac{\left(\dfrac{1}{50}+1\right)+\left(\dfrac{2}{49}+1\right)+...+\left(\dfrac{49}{2}+1\right)+1}{7\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{51}\right)}\)
\(=\dfrac{\dfrac{51}{50}+\dfrac{51}{49}+...+\dfrac{51}{2}+\dfrac{51}{51}}{7\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{51}\right)}=\dfrac{51\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{51}\right)}{7\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{51}\right)}\)
\(=\dfrac{51}{7}\)
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