Question

Find the coefficient of 'x³ in the expansion of (3x + 4)⁶ pls answer fast with every step​

Answer 1

EXPLANATION.

Coefficient of x³ in the expansion.

Expansion = (3x + 4)⁶.

As we know that,

Formula of :

Binomial theorem for positive integral index.

⇒ (x + a)ⁿ = ⁿC₀xⁿa⁰ + ⁿC₁xⁿ⁻¹a + ⁿC₂xⁿ⁻²a² + . . . . . + ⁿCₙxaⁿ.

Using this formula in the equation, we get.

⇒ (3x + 4)⁶ = $\sf ^{6}C_{r} (3x)^{6 - r} (4)^{r}$.

$\sf \implies ^{6}C_{r} (3)^{6 - r} (x)^{6 - r} (4)^{r} .$

$\sf \implies (x)^{6 - r} = (x)^{3} .$

$\sf \implies 6 - r = 3.$

$\sf \implies r = 3.$

Put the values of r = 3 in the expansions, we get.

⇒ ⁶C₃(3)⁶⁻³ (x)⁶⁻³(4)³.

⇒ ⁶C₃(3)³(4)³(x)³.

⇒ ⁶C₃ (27) x (64) (x)³.

⇒ [6!/3! x 3!] (27)(64) (x)³.

⇒ [6 x 5 x 4 x 3!/3 x 2 x 1 x 3!] (27)(64) (x)³.

⇒ [120/6] (27)(64)(x)³.

⇒ 20 x 27 x 64 x (x)³.

⇒ 34560 (x)³.

                                                                                                                       

MORE INFORMATION.

Number of terms in the expansions of (x + y + z)ⁿ.

ⁿ⁺²C₂ = (n + 1)(n + 2)/2.

Number of terms in the expansions of (x₁ + x₂ + x₃ + . . . . . + x_{k})ⁿ are,

$\sf ^{n + k - 1}C_{k - 1}$  when x₁, x₂, x₃ . . . . . .x_{k} all are different and can not be solved.

Answer 2

Step-by-step explanation:

given :

• Find the coefficient of 'x³ in the expansion of (3x + 4)⁶ pls answer fast with every step

to find :

• Find the coefficient of 'x

solution :

• (3x + 4)6 = 6C₁ (3x)6—r(4)r .

• 6C, (3)6-(x)6-r(4)'.

• (x) 6-r = (x)³.

• 6- r = 3.

• ⇒ r = 3.

• 6C₂(3)6-3 (x)6-3(4)³.

• 6C₂(3)³(4)³(x)³.

• °C3 (27) x (64) (x)³.

• [6!/3! x 3!] (27)(64) (x)³.

• [6 x 5 x 4 x 3!/3 x 2 x 1 x 3!] (27)(64) (x)³.

• [120/6] (27)(64)(x)³.

• 20 x 27 x 64 x (x)³.

• 34560 (x)³.