Question

WILL NAME BRAINLIEST PLEASE HELP ASAP!

Answer 1

Important:  to denote exponentiation use " ^ ":

(x + y)1 = ___ x + ___ y    NO
(x + y)^1 = ___ x + ___ y   YES

(x+y)^1 = 1x + 1y

(x+y)^2 = 1x + 2xy + y^2

(x+y)^3 = 1x^3 + 3x^2*y + 3x*y^2 + y^3

and so on.   Look up "Pascal's Triangle" if you want more info on this pattern.

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(x + y)4 = ___ x4 + ___ x3y + ___ x2y2 + ___ xy3 + ___ y4    NO

(x + y)^4 = ___ x^4 + ___ x^3y + ___ x^2y^2 + ___ xy^3 + ___ y^4    YES

 (x+y)^4 =  1x^4 + 4x^3*y + 6x^2*y^2 + 4x*y^3 + y^4

Answer 2

Answer:

$a_0=1, a_1=1$

$a_2=1, a_3=2,a_4=1$

$a_5=1,a_6= 3,a_7=3,a_8=1$

$a_9=1,a_1_0=4,a_1_1=6,a_1_2=4,a_1_3=1$.

Step-by-step explanation:

$(x+y)^1=\binom{1}{0}x+ \binom{1}{1}y$

$\binom{n}{r}=\frac{n!}{r!(n-r)!}$

 Expansion:$(x+y)^n=a_0x^n+a_1x^{n-1}y+a_2x^{n-2}y^2+...............a_ny^n$

then we get

$(x+y)^1= x+y$

Therefore, we get $a_0=1,a_1=1$

$(x+y)^2=\binom{2}{0}x^2+\binom{2}{1}xy+\binom{2}{2}y^2$

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

Therefore,$a_2=1,a_3=2,a_4=1$

$(x+y)^3=\binom{3}{0}x^3+\binom{3}{1}x^2y+\binom{3}{2}xy^2+\binom{3}{3}y^3$

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

Therefore, $a-5=1,a_6=3,a_7=3,a_8=1$

$(x+y)^4=\binom{4}{0}x^4+\binom{4}{1}x^3y+\binom{4}{2}x^2y^2+\binom{4}{3}xy^3+\binom{4}{4}y^4$

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

Therefore, $a_9=1,a_1_0=4,a_1_1=6,a_1_2=4,a_1_3=1$.