Express as a continued fraction the series 1 a 0 - x a 0 a 1 - x 2 a 0 a 1 a 2 - - -1 n x n a 0 a 1 a 2 - a n-

Put 1 a _ n x a _ n a _ n+1 = 1 a _ n + y _ n ;then ( a _ n + y _ n ) ( a _ n+1 x ) = a _ n a _ n+1 ; #160; #160; #160; #160; ...

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Byju's Answer

Standard IX

Mathematics

Method of Difference

Express as a ...

Question

Express as a continued fraction the series
$\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} - \dots + {(- 1)}^{n} \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}}$ .

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Solution

Put $\frac{1}{a_{n}} - \frac{x}{a_{n} a_{n + 1}} = \frac{1}{a_{n} + y_{n}}$ ;
then $(a_{n} + y_{n}) (a_{n + 1} - x) = a_{n} a_{n + 1}$ ;
$∴ y_{n} = \frac{a_{n} x}{a_{n + 1} - x}$ .
Hence $\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} = \frac{1}{a_{0} + y_{0}} = \frac{1}{a_{0} +} \frac{a_{0} x}{a_{1} - x}$ .
Again, $\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} = \frac{1}{a_{0}} - \frac{x}{a_{0}} (\frac{1}{a_{1}} - \frac{x}{a_{1} a_{2}}) = \frac{1}{a_{0}} - \frac{x}{a_{0} (a_{1} + y_{1})}$
$= \frac{1}{a_{0} +} \frac{a_{0} x}{a_{1} + y_{1} - x}$
$= \frac{1}{a_{0} +} \frac{a_{0} x}{a_{1} - x +} \frac{a_{1} x}{a_{2} - x}$ ;
and generally $\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} - \dots + {(- 1)}^{n} \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}}$
$= \frac{1}{a_{0} +} \frac{a_{0} x}{a_{1} - x +} \frac{a_{1} x}{a_{2} - x +} \dots \frac{a_{n - 1} x}{a_{n} - x}$ .

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\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}

If $(1 + a x + b x^{2})^{4} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{8} x^{8}; a, b, a_{0}, a_{1} \dots a_{8} ϵ R$ and are such that $a_{0} + a_{1} + a_{2} \neq 0$ and $∣ ∣ ∣ ∣ \begin{matrix} a_{0} & a_{1} & a_{2} a_{1} & a_{2} & a_{0} a_{2} & a_{0} & a_{1} \end{matrix} ∣ ∣ ∣ ∣ = 0$ , then