Question

find the term independent of x in the expansion of (2x - 1/3x^2)^9​

Answer 1

EXPLANATION.

⇒ (2x - 1/3x²)⁹.

As we know that,

We can write equation as,

$\sf \implies T_{r + 1} \ = \ ^{9}C_{r} (2x)^{9 - r} . \bigg(\dfrac{-1}{3x^{2} } \bigg)^{r}$

$\sf \implies T_{r + 1} \ = (-1)^{r} \ ^{9} C_{r} (2^{9 - r} ).(x^{9 - r} ) . \bigg(\dfrac{1}{3^{r}.x^{2r} } \bigg)$

$\sf \implies T_{r + 1} \ = (-1)^{r} \ ^{9} C_{r} (2^{9 - r} ).(x^{9 - r} ) . (3^{-r} ).(x^{-2r} )$

$\sf \implies T_{r + 1} \ = (-1)^{r} \ ^{9} C_{r} (2^{9 - r} ).(x^{9 - 3r} ) .(3^{-r} )$

$\sf \implies 9 - 3r = 0.$

$\sf \implies 9 = 3r.$

$\sf \implies t = 3.$

$\sf \implies T_{r + 1} = 3 + 1 = 4th \ term.$

Put the value of r = 3 in equation, we get.

$\sf \implies (-1)^{3} \ ^{9} C_{3} (2^{9 - 3} ).(3)^{-3}$

$\sf \implies (-1).^{9} C_{3} (2)^{6} .(3)^{-3}$

$\sf \implies (-1) \ ^{9} C_{3} (64)(-27).$

$\sf \implies ^{9} C_{3} (1728). = \ \dfrac{9 \times 8 \times 7 \times 6!}{3! \times 6!} \times 1728$

$\sf \implies \dfrac{9 \times 8 \times 7 }{3 \times 2 \times 1} \times 1728$

$\sf \implies \dfrac{504 \times 1728}{6} \ = 145,152$

                                                                                                                       

MORE INFORMATION.

Some important expansions.

(1) = (1 - x)⁻¹ = 1 + x + x² + x³ + . . . . . + x^(r) + . . .

General term = T(r + 1) = x^(r).

(2) = (1 + x)⁻¹ = 1 - x + x² - x³ + . . . . . (-x)^(r) + . . .

General term = T(r + 1) = (-x)^(r).

(3) = (1 - x)⁻² = 1 + 2x + 3x² + 4x³ + . . . . . + (r + 1)x^(r) + . . .

General term = T(r + 1) = (r + 1)x^(r).

(4) = (1 + x)⁻² = 1 - 2x + 3x² - 4x³ + . . . . + (r + 1)(-x)^(r) + . . .  .

General term = T(r + 1) = (r + 1)(-x)^(r).