Math, asked by mk02032004, 6 months ago

Find the sum of the series
$(1 + x) + (1 + x + {x}^{2}) + (1 + x + {x}^{2} + {x}^{3} ) + ............$

Answers

Answered by Anonymous

$\bf \large \bold{Hola! }$

$\sf(1 + x) + (1 + x + {x}^{2} ) + (1 + x + {x}^{2} + {x}^{3} ) + ...$

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→ Sum of the series

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$\sf \: S_n=a+ar + a {r}^{2} + ... = \frac{a( {r}^{n} - 1) }{(r - 1)} \\$

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$\sf \: (1 + x) + (1 + x + {x}^{2}) + (1 + x + {x}^{2} + {x}^{3} ) + ...$

$\implies \sf \: \frac{1 \times (x - 1)}{(x - 1)} + \frac{1( {x}^{2} - 1) }{(x - 1)} + \frac{1( {x}^{3} - 1) }{(x - 1)} + ... \\$

$\implies \sf \: \frac{1}{(x - 1)} \{x - 1 + {x}^{2} - 1 + {x}^{3} - 1 + ... \} \\$

$\implies \: \sf \frac{1}{(x - 1)} \{(x + {x}^{2} + {x}^{3} + ... + {x}^{n}) - n \} \\$

$\implies \sf \: \frac{1}{(x - 1)} \{ \frac{x( {x}^{n} - 1)}{(x - 1)} - n \} \\$

$\implies \sf \: { \underline{\boxed{ \sf\ \frac{( {x}^{n + 1} - x)}{(x - 1)^{2} } - \frac{n}{x-1} }}}\\$

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