Question

Mention the condition for validity of the expansion and expand (2-3x)^-3 as far as the term containing x^4.​

Answer 1

Answer :-

${(2 - 3x)}^{ - 3} \: can \: be \: written \: as \: -$

$= {2}^{ - 3} {[1 - \frac{3}{2} x]}^{ - 3}$

$= \frac{1}{8} {[1 - \frac{3}{2} x]}^{ - 3}$

The expansion is valid when

$| \frac{3}{2}x | < 1 \: \: i.e \: when \: |x| < \frac{2}{3}$

The required expansion is :-

${(2 - 3x)}^{ - 3} = \frac{1}{8} {[ 1 - \frac{3}{2} x]}^{ - 3}$

$= \frac{1}{8} [1 + ( - 3)( - \frac{3}{2} x) + \frac{( - 3)( - 4)}{2!} {( - \frac{3}{2} x)}^{2} \\ + \frac{( - 3)( - 4)( - 5)}{3!}( { - \frac{3}{2}x })^{3} + \frac{( - 3)( - 4)( - 5)( - 6)}{4!} ( - { \frac{ 3}{2} x)}^{4} ]$

$= \frac{1}{8} [1 + \frac{9x}{2} + \frac{ {27x}^{2} }{2} + \frac{ {135x}^{3} }{4} + \frac{ {1215x}^{4} }{16} + - - ]$

$= [ \frac{1}{8} + \frac{9x}{16} + \frac{ {27x}^{2} }{16} + \frac{ {135x}^{3} }{32} + \frac{ {1215x}^{4} }{128} + - - - ]$

So , the Required answer is ,

$= [ \frac{1}{8} + \frac{9x}{16} + \frac{ {27x}^{2} }{16} + \frac{ {135x}^{3} }{32} + \frac{ {1215x}^{4} }{128} + - - - ]$