Question Number 86560 by naka3546 last updated on 29/Mar/20
$${x}\:\:=\:\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{15}}{\mathrm{32}}\:+\:\frac{\mathrm{105}}{\mathrm{384}}\:+\:… \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{1}\:\:=\:\:? \\ $$
Commented by redmiiuser last updated on 29/Mar/20
$${ans}\:{is}\:\mathrm{7}. \\ $$
Commented by redmiiuser last updated on 29/Mar/20
$${pls}\:{check}\:{the}\:{ans}\:{sir}. \\ $$
Commented by Prithwish Sen 1 last updated on 29/Mar/20
$$\mathrm{the}\:\mathrm{series}\:\mathrm{is} \\ $$$$\left(\mathrm{1}−\mathrm{a}\right)^{−\:\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{where}\:\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\: \\ $$$$\therefore\:\boldsymbol{\mathrm{x}}\:=\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\:\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}\:=\:\mathrm{8}−\mathrm{1}\:=\:\mathrm{7} \\ $$
Commented by naka3546 last updated on 29/Mar/20
$${thank}\:\:{you},\:{sir} \\ $$