Show that 1 u 0 - 1 u 1 - 1 u 2 - 1 u 3 - - -1 n 1 u n - 1 u 0 - u 0 2 u 1 - u 0 - u 1 2 u 2 - u 1 - - u n-1 2 u n - u n-1-

Put, 1 u _ r 1 u _ r+1 = 1 u _ r + x _ r ( u _ r+1 u _ r ) ( u _ r + x _ r ) = u _ r u _ r+1 ; So that x _ r = u _ r ^ 2 u _ r+1 u ...

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Arithmetic Mean

Show that 1 ...

Question

Show that
$\frac{1}{u_{0}} - \frac{1}{u_{1}} + \frac{1}{u_{2}} - \frac{1}{u_{3}} + \dots + {(- 1)}^{n} \frac{1}{u_{n}} = \frac{1}{u_{0} +} \frac{u_{0}^{2}}{u_{1} - u_{0} +} \frac{u_{1}^{2}}{u_{2} - u_{1} +} \dots \frac{u_{n - 1}^{2}}{u_{n} - u_{n - 1}}$ .

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Similar questions

u_{1} + u_{2} + u_{3} + u_{4} + . . . . . .

lim n \to \infty \sqrt[n]{u_{n}}

< 1

> 1

u_{1}

u_{2}

u_{n + 2} + u_{n} f o r n \geq 1

u_{n} = \frac{1}{\sqrt{(5)}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]

n \geq 1

u_{n} = 2 cos n θ

u_{1} u_{n} - u_{n - 1}

\frac{u_{n}}{u_{n + 1}} = \frac{n^{k} + A n^{k - 1} + B n^{k - 2} + C n^{k - 3} + . . . . .}{n^{k} + a n^{k - 1} + b n^{k - 2} + c n^{k - 3} + . . . . .}

k

u_{1} + u_{2} + u_{3} + . . . . . .

A - a - 1

A - a - 1

u_{n} = {sin}^{n} θ + {cos}^{n} θ,

\frac{u_{5} - u_{7}}{u_{3} - u_{5}} = \frac{u_{3}}{u_{1}}

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