Determine-the-term-independent-of-x-in-the-expansion-x-1-x-2-3-x-1-3-1-x-1-x-x-1-2-10-

Question Number 137943 by john_santu last updated on 08/Apr/21

D e t e r m i n e t h e t e r m i n d e p e n d e n t

o f x i n t h e e x p a n s i o n

{(\frac{x + 1}{x^{2 / 3} - x^{1 / 3} + 1} - \frac{x - 1}{x - x^{1 / 2}})}^{10} .

Answered by EDWIN88 last updated on 08/Apr/21

{[\frac{x + 1}{x^{2 / 3} - x^{1 / 3} + 1} - \frac{x - 1}{x - x^{1 / 2}}]}^{10} =

{[\frac{(x^{1 / 3} + 1) (x^{2 / 3} - x^{1 / 3} + 1)}{x^{2 / 3} - x^{1 / 3} + 1} - \frac{(x^{1 / 2} - 1) (x^{1 / 2} + 1)}{x^{1 / 2} (x^{1 / 2} - 1)}]}^{10} =

{[(x^{1 / 3} + 1) - \frac{x^{1 / 2} + 1}{x^{1 / 2}}]}^{10} =

{[x^{1 / 3} - x^{- 1 / 2}]}^{10} = \underset{k = 0}{\sum^{10}} C (10, k) x^{k / 3} {(- x^{- 1 / 2})}^{10 - k}

t h e t e r m i n d e p e n d e n t o f x i t m u s t b e

f r o m x^{\frac{k}{3} + \frac{k}{2} - 5} = x^{0}, \frac{5 k}{6} = 5 \Rightarrow k = 6

t h e n T_{6} = (\begin{matrix} 10 \\ 6 \end{matrix}) {(x^{1 / 3})}^{6} {(- x^{- 1 / 2})}^{4}

T_{6} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} {(- 1)}^{4} = 210

Facebook Tweet Pin