Question

See photo please, I don’t know what it is.

Answer 1

Question:

• Find the Coefficient of ${x}^{3}$ in the expansion of ${(2+3x)}^{5}$, giving your answer as simply as possible.

Answer:

$\large\bold\red{1080}$

Step-by-step explanation:

Given,

${(2 + 3x)}^{5}$

We know that,

In the expansion of ${(a+x)}^{n}$,

General term , (r+1)th term is given by,

$\large \boxed{ \bold \purple{T_{r+1}= ^n{C}_{r} {a}^{r} {x}^{n - r} }}$

Now,

In the given question,

We have ,

• n = 5
• a = 2
• x = 3x

Therefore,

General term, (r+1)th term is given as,

$\large \boxed{ \bold{T_{r+1}=^5C_{r} {2}^{r} {(3x)}^{5 - r} }}$

Further solving,

We get,

$= > T_{r+1}=\:^5C_{r} \times {2}^{r} \times {3}^{5 - r} \times {x}^{5 - r} \\ \\ = > \bold{ T_{r+1}=\:^5C_{r} {2}^{r} {3}^{5-r}{x}^{ 5- r} }$

Now,

We have to find the power of x = 3.

Therefore,

We get,

$= > 5 - r = 3 \\ \\ = > r = 5 - 3 \\ \\ = > r = 2$

Therefore,

Coefficient of ${x}^{3}$ is,

$= \: ^5 C_{2} \times {2}^{2} \times{3}^{3} \\ \\ = \frac{5!}{2!(5 - 2)!} \times {2}^{2} {3}^{3}\\ \\ = \frac{5!}{3!2!} \times {2}^{2}{3}^{3} \\ \\ = \frac{5 \times 4 \times \cancel{3!}}{2 \times \cancel{3!}} \times {2}^{2} {3}^{3}\\ \\ = 5 \times {2}^{2} \times {2}^{2- 1} {3}^{3}\\ \\ = 5 \times {2}^{1 + 2} {3}^{3}\\ \\ = 5 \times {2}^{3}{3}^{3} \\ \\ = 5 \times 8 \times 27 \\ \\ = 1080$

Hence,

$\large\bold{1080}$ is the required coefficient.

Answer 2

Question :---

• Find the coefficient of x³ in the expansion of (2+3x)^5 ?

Concept used :----

• i will solve this problem by pascal ∆ method , because by binomial theraom calculation is little bit lengthy .

Solution :----

(2+3x)^5

Here, coefficinent of x³ will be = 5(C)3 × (3x)³ × (2)²

5(C)3 = 5!/3!×2! = 5×4/2 = 10

so,

→ 10 × (3x)³ × 4

→ 40 × 27x³

→ 1080 x³

Hence , coefficient of x³ is 1080 (Ans) ...

(Hope this Helps you in Easiest way)