tính tổng S=1+3+3^2+3^3+...+3^10
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\(S=3^1+3^2+3^3+.....+3^{100}\) \(=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=120+3^5.\left(3^1+3^2+3^3+3^4\right)+....+3^{97}.\left(3^1+3^2+3^3+3^4\right)\)
\(=1.120+3^5.120+...+3^{97}.120\)
\(=\left(1+3^5+...+3^{97}\right).120\)
\(\Rightarrow S⋮120\)
Vậy ........
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\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times.....\times\left(1-\frac{1}{99}\right)\times\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{98}{99}\times\frac{99}{100}\)
\(=\frac{1}{100}\)
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S=1+3+3^2+3^3+...+3^10
3.S=3+3^2+3^3+3^4+...+3^11
3.S-S=(3+3^2+3^3+3^4+...+3^10)-(1+3+3^2+3^3+...+3^10
3.S-S=3+3^2+3^3+3^4+...+3^11-1-3-3^2-3^3-...-3^10
S=3^11-1