x-x-1-2-x-2-x-2-2-x-3-x-3-2-x-10-x-10-2-

Question Number 85832 by jagoll last updated on 25/Mar/20

{(x + x^{- 1})}^{2} + {(x^{2} + x^{- 2})}^{2} + {(x^{3} + x^{- 3})}^{2}

+ \dots + {(x^{10} + x^{- 10})}^{2} =

Commented by mathmax by abdo last updated on 25/Mar/20

l e t A_{n} (x) = {(x + x^{- 1})}^{2} + {(x^{2} + x^{- 2})}^{2} + {(x^{3} + x^{- 3})}^{2} + \dots + {(x^{n} + x^{- n})}^{2}

\Rightarrow A_{n} = x^{2} + \frac{1}{x^{2}} + 2 + x^{4} + \frac{1}{x^{4}} + 2 + \dots . x^{2 n} + \frac{1}{x^{2 n}} + 2

= x^{2} + x^{4} + \dots . + x^{2 n} + {(\frac{1}{x})}^{2} + {(\frac{1}{x})}^{4} + \dots . + {(\frac{1}{x})}^{2 n} + 2 n

= w (x) + w (\frac{1}{x}) + 2 n w i t h w (x) = x^{2} + x^{4} + \dots x^{2 n}

= x^{2} (1 + x^{2} + \dots + {(x^{2})}^{n - 1}) = x^{2} \times \frac{1 - {(x^{2})}^{n}}{1 - x^{2}} = \frac{x^{2}}{1 - x^{2}} (1 - x^{2 n})

w (\frac{1}{x}) = \frac{1}{x^{2} (1 - \frac{1}{x^{2}})} (1 - \frac{1}{x^{2 n}}) = \frac{1}{x^{2}} (1 - \frac{1}{x^{2 n}}) = \frac{x^{2 n} - 1}{x^{2 n + 2}} \Rightarrow

A_{n} (x) = \frac{x^{2} - x^{2 n + 2}}{1 - x^{2}} + \frac{x^{2 n} - 1}{x^{2 n + 2}} + 2 n (x \neq \bar{+} 1)

n = 10 \Rightarrow A_{10} (x) = \frac{x^{2} - x^{22}}{1 - x^{2}} + \frac{x^{20} - 1}{x^{22}} + 20

Answered by TANMAY PANACEA. last updated on 25/Mar/20

T_{r} = {(x^{r} + x^{- r})}^{2} = x^{2 r} + 2 + \frac{1}{x^{2 r}}

T_{1} + T_{2} + T_{3} + \dots + T_{10}

= (x^{2} + x^{4} + x^{6} + . . + x^{20}) + 2 \times 10 + (\frac{1}{x^{2}} + \frac{1}{x^{4}} + \frac{1}{x^{6}} + . . + \frac{1}{x^{20}})

s = \frac{a (1 - r^{n})}{1 - r}

= \frac{x^{2} (1 - x^{20})}{1 - x^{2}} + 20 + \frac{\frac{1}{x^{2}} (1 - \frac{1}{x^{20}})}{1 - \frac{1}{x^{2}}}

Commented by jagoll last updated on 25/Mar/20

thank you sir

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