Respuesta :
QUESTION 1
We want to expand [tex](x-2)^6[/tex].
We apply the binomial theorem which is given by the formula
[tex](a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+...+^nC_na^{n-n}b^n[/tex].
By comparison,
[tex]a=x,b=-2,n=6[/tex].
We substitute all these values to obtain,
[tex](x-2)^6=^6C_0x^6(-2)^0+^6C_1x^{6-1}(-2)^1+^6C_2x^{6-2}(-2)^2+^6C_3x^{6-3}(-2)^3+^6C_4x^{6-4}(-2)^4+^6C_5x^{6-5}(-2)^5+^6C_6x^{6-6}(-2)^6[/tex].
We now simplify to obtain,
[tex](x-2)^6=^nC_0x^6(-2)^0+^6C_1x^{5}(-2)^1+^6C_2x^{4}(-2)^2+^6C_3x^{3}(-2)^3+^6C_4x^{2}(-2)^4+^6C_5x^{1}(-2)^5+^6C_6x^{0}(-2)^6[/tex].
This gives,
[tex](x-2)^6=x^6-12x^{5}+60x^{4}-160x^{3}(-2)^3+240x^{2}-1925x+64[/tex].
Ans:C
QUESTION 2
We want to expand
[tex](x+2y)^4[/tex].
We apply the binomial theorem to obtain,
[tex](x+2y)^4=^4C_0x^4(2y)^0+^4C_1x^{4-1}(2y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(2y)^3+^4C_4x^{4-4}(2y)^4[/tex].
We simplify to get,
[tex](x+2y)^4=x^4(2y)^0+4x^{3}(2y)^1+6x^{2}(2y)^2+4x^{1}(2y)^3+x^{0}(2y)^4[/tex].
We simplify further to obtain,
[tex](x+2y)^4=x^4+8x^{3}y+24x^{2}y^2+32x^{1}y^3+16y^4[/tex]
Ans:B
QUESTION 3
We want to find the number of terms in the binomial expansion,
[tex](a+b)^{20}[/tex].
In the above expression, [tex]n=20[/tex].
The number of terms in a binomial expression is [tex](n+1)=20+1=21[/tex].
Therefore there are 21 terms in the binomial expansion.
Ans:C
QUESTION 4
We want to expand
[tex](x-y)^4[/tex].
We apply the binomial theorem to obtain,
[tex](x-y)^4=^4C_0x^4(-y)^0+^4C_1x^{4-1}(-y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(-y)^3+^4C_4x^{4-4}(-y)^4[/tex].
We simplify to get,
[tex](x+2y)^4=^x^4(-y)^0+4x^{3}(-y)^1+6x^{2}(-y)^2+4x^{1}(-y)^3+x^{0}(-y)^4[/tex].
We simplify further to obtain,
[tex](x+2y)^4=x^4-4x^{3}y+6x^{2}y^2-4x^{1}y^3+y^4[/tex]
Ans: C
QUESTION 5
We want to expand [tex](5a+b)^5[/tex]
We apply the binomial theorem to obtain,
[tex](5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{5-1}(b)^1+^5C_2(5a)^{5-2}(b)^2+^5C_3(5a)^{5-3}(b)^3+^5C_4(5a)^{5-4}(b)^4+^5C_5(5a)^{5-5}(b)^5[/tex].
We simplify to obtain,
[tex](5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{4}(b)^1+^5C_2(5a)^{3}(b)^2+^5C_3(5a)^{2}(b)^3+^5C_4(5a)^{1}(b)^4+^5C_5(5a)^{0}(b)^5[/tex].
This finally gives us,
[tex](5a+b)^5=3125a^5+3125a^{4}b+1250a^{3}b^2+^250a^{2}(b)^3+25a(b)^4+b^5[/tex].
Ans:B
QUESTION 6
We want to expand [tex](x+2y)^5[/tex].
We apply the binomial theorem to obtain,
[tex](x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{5-1}(2y)^1+^5C_2(x)^{5-2}(2y)^2+^5C_3(x)^{5-3}(2y)^3+^5C_4(x)^{5-4}(2y)^4+^5C_5(x)^{5-5}(2y)^5[/tex].
We simplify to get,
[tex](x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{4}(2y)^1+^5C_2(x)^{3}(2y)^2+^5C_3(x)^{2}(2y)^3+^5C_4(x)^{1}(2y)^4+^5C_5(x)^{0}(2y)^5[/tex].
This will give us,
[tex](x+2y)^5=x^5+^10(x)^{4}y+40(x)^{3}y^2+80(x)^{2}y^3+80(x)y^4+32y^5[/tex].
Ans:A
QUESTION 7
We want to find the 6th term of [tex](a-y)^7[/tex].
The nth term is given by the formula,
[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].
Where [tex]r=5,n=7,b=-y[/tex]
We substitute to obtain,
[tex]T_{5+1}=^7C_5a^{7-5}(-y)^5[/tex].
[tex]T_{6}=-21a^{2}y^5[/tex].
Ans:D
QUESTION 8.
We want to find the 6th term of [tex](2x-3y)^{11}[/tex]
The nth term is given by the formula,
[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].
Where [tex]r=5,n=11,a=2x,b=-3y[/tex]
We substitute to obtain,
[tex]T_{5+1}=^{11}C_5(2x)^{11-5}(-3y)^5[/tex].
[tex]T_{6}=-7,185,024x^{6}y^5[/tex].
Ans:D
QUESTION 9
We want to find the 6th term of [tex](x+y)^8[/tex].
The nth term is given by the formula,
[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].
Where [tex]r=5,n=8,a=x,b=y[/tex]
We substitute to obtain,
[tex]T_{5+1}=^8C_5(x)^{8-5}(y)^5[/tex].
[tex]T_{6}=56a^{3}y^5[/tex].
Ans: A
We want to find the 7th term of [tex](x+4)^8[/tex].
The nth term is given by the formula,
[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].
Where [tex]r=6,n=8,a=x,b=4[/tex]
We substitute to obtain,
[tex]T_{6+1}=^8C_5(x)^{8-6}(4)^6[/tex].
[tex]T_{7}=114688x^{2}[/tex].
Ans:A
