Question

Given that (1 - x) (1 + x + x² + x3 + x4) is 31/32and x is a rational number, what is1 + x + x2 + x3 + x4 + x5?​

Answer 1

Answer:

$\frac{63}{32}$

Step-by-step explanation:

Given:

$(1 - x)(1 + x + {x}^{2} + {x}^{3} + {x}^{4} ) = \frac{31}{32}$

$(1 + x + {x}^{2} + {x}^{3} + {x}^{4} ) - (x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} ) = \frac{31}{32}$

$1 - {x}^{5} = \frac{31}{32}$

$1 - \frac{31}{32} = {x}^{5}$

$\frac{1}{32} = {x}^{5}$

$= > x = \frac{1}{2}$

Hence we got value of x

From now,

Required value is

$1 + x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} = 1 + \frac{1}{2} + { \frac{1}{2} }^{2} + ....... + { \frac{1}{2} }^{5}$

From sum of n terms of a GP definition

$\frac{1 - { \frac{1}{2} }^{6} }{1 - \frac{1}{2} }$

$\frac{ {2}^{6} - 1 }{ {2}^{5} }$

$\frac{64 - 1}{32} = \frac{63}{32}$

Done!

Thank you!