LT

\(\frac{1}{1-x}\)+\(\frac{1}{1+x}\)+\(\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)

VV
6 tháng 11 2016 lúc 8:59

Theo đầu bài ta có:
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(1+x\right)+\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(2+2x^2\right)+\left(2-2x^2\right)}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(4+4x^4\right)+\left(4-4x^4\right)}{1-x^8}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{1}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{1}{1+x^{16}}\)
\(=\frac{\left(8+8x^8\right)+\left(8-8x^8\right)}{1-x^{16}}+\frac{1}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{1}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)+\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{\left(16+16x^{16}\right)+\left(1-x^{16}\right)}{1-x^{32}}\)
\(=\frac{17+15x^{16}}{1-x^{32}}\)

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