Question

Applying the Binomial Theorem:

Answer 1

Answer:

$(x+y)^4=x^4+x^3y+x^2y^2+xy^3+y^4$

Step-by-step explanation:

We have to expand

$(x+y)^4$

we can use rth term formula of binomial theorem

$(a+b)^n$

$T_r_+_1=a^{n-r}(b)^{r}$

$(x+y)^4$

we can see that

a=x

b=y

n=4

Part-A:

we can use formula

$T_r_+_1=a^{n-r}(b)^{r}$

and we get

$T_1=x^{4-0}(y)^{0}$

$T_1=x^{4}$

Part-B:

we can use formula

$T_r_+_1=a^{n-r}(b)^{r}$

and we get

$T_2=x^{4-1}(y)^{1}$

$T_2=x^{3}y$

Part-C:

we can use formula

$T_r_+_1=a^{n-r}(b)^{r}$

and we get

$T_3=x^{4-2}(y)^{2}$

$T_3=x^{2}y^2$

Part-D:

we can use formula

$T_r_+_1=a^{n-r}(b)^{r}$

and we get

$T_4=x^{4-3}(y)^{3}$

$T_4=xy^3$

Part-E:

we can use formula

$T_r_+_1=a^{n-r}(b)^{r}$

and we get

$T_5=x^{4-4}(y)^{4}$

$T_5=y^4$

Part-F:

now, we can combine them

and we get

$(x+y)^4=x^4+x^3y+x^2y^2+xy^3+y^4$

Answer 2

Answer:

Part A: x⁴

Part B: 4x³y¹

Part C: 6x²y²

Part D: 4x¹y³

Part E: y⁴

Part F: x⁴ + 4x³y¹ + 6x²y² + 4x¹y³ + y⁴

Step-by-step explanation:

Part A:

The  first term of the expansion (x + y)⁴

the first co-efficient = ⁴C₀ = 1

term = x⁴⁻⁰y⁰

First term = 1x⁴y⁰ = x⁴

Part B:

the second co-efficient = ⁴C₁ =  4

term = x⁴⁻¹y¹ = x³y¹

Second term = 4x³y¹

Part C:

the third co-efficient = ⁴C₂ = 6

term = x⁴⁻²y² = x³y

Third term = 6x²y²

Part D:

the fourth co-efficient = ⁴C₃ = 4

term = x⁴⁻³y³ = x³y³

Fourth term = 4x¹y³

Part D:

the fifth co-efficient = ⁴C₄ =  1

term = x⁴⁻⁴y⁴ = x⁰y⁴

Fifth term = y⁴

Part E:

The answer is the combination of part A, B, C, D, and F.

(x + y)⁴ = x⁴ + 4x³y¹ + 6x²y² + 4x¹y³ + y⁴