Find the negative of the coefficient of $x^{3}$ in the expansion of ${(1 - x + x^{2})}^{6}$ .

Open in App

Solution

${(1 - x + x^{2})}^{6} = {1 - x (1 - x)}^{6}$
$=^{6} C_{0} -^{6} C_{1} x (1 - x) +^{6} C_{2} x^{2} {(1 - x)}^{2} -^{6} C_{3} x^{3} {(1 - x)}^{3} + . . .$ to $7$ terms
$=^{6} C_{0} -^{6} C_{1} x (1 - x) +^{6} C_{2} x^{2} (1 - 2 x + x^{2}) -^{6} C_{3} x^{3} (1 - 3 x + 3 x^{2} - x^{3}) + . . .$ to $7$ terms
$∴$ Coefficients of $x^{3} = - {2.}^{6} C_{2} -^{6} C_{3},$ $($ collecting coefficients of $x^{3}$ from each term $)$
$= - 2. \frac{6!}{2! 4!} - \frac{6!}{3! 3!} = - 50$

Suggest Corrections

Similar questions

Find the coefficient of $x^{3}$ in the expansion of $(1 + x + x^{2})^{5}$

x^{n}

\frac{2 + x + x^{2}}{(1 + x)^{3}}

x

(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . + \frac{x^{n}}{n!})

(1 + x + x^{2} + x^{3} + . . . . .

\infty)^{3}

x^{3}

Expand ${(x^{3} - \frac{2}{x^{2}})}^{6}$ using binomial expansion.

Join BYJU'S Learning Program