Question

Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion of (3a + 4b)8, which of the following are possible variable terms? Explain your reasoning. a2b3; a5b3; ab8; b8; a4b4; a8; ab7; a6b5

Answer 1

$\bf (2x+3)^5\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&&(2x)^5(+3)^0\\ 2&+5&(2x)^4(+3)^1\\ 3&+10&(2x)^3(+3)^2\\ 4&+10&(2x)^2(+3)^3\\ 5&+5&(2x)^1(+3)^4\\ 6&+1&(2x)^0(+3)^5 \end{array}$

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


$\bf (2x-3y)^4\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&&(2x)^4(-3y)^0\\ 2&+4&(2x)^3(-3y)^1\\ 3&+6&(2x)^2(-3y)^2\\ 4&+4&(2x)^1(-3y)^3\\ 5&+1&(2x)^0(-3y)^4 \end{array}$


$\bf (3a+4b)^8\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&&(3a)^8(+4b)^0\\ 2&+8&(3a)^7(+4b)^1\\ 3&+28&(3a)^6(+4b)^2\\ 4&+56&(3a)^5(+4b)^3\\ 5&+70&(3a)^4(+4b)^4\\ 6&+56&(3a)^3(+4b)^5\\ 7&+28&(3a)^2(+4b)^6\\ 8&+8&(3a)^1(+4b)^7\\ 9&+1&(3a)^0(+4b)^8 \end{array}$

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶

Answer 2

Answer:

The possible terms are $a^5b^3,b^8,a^4b^4,a^8,ab^7$.

Step-by-step explanation:

Binomial Theorem:

$(a+b)^n=^nC_0a^n+^nC_1a^{n-1}b+...+^nC_{n-1}a^1b^{n-1}+^nC_nb^n$

Using Binomial Theorem, we get

$(x + 2)^6=^6C_0a^n+^6C_1a^{5}b+...+^6C_{5}a^1b^{5}+^5C_5b^5$

$(x + 2)^6=x^6 + 12 x^5 + 60 x^4 + 160 x^3 + 240 x^2 + 192 x + 64$

Similarly,

$(x-4)^4=x^4 - 16 x^3 + 96 x^2 - 256 x + 256$

$(2x+3)^5=32 x^5 + 240 x^4 + 720 x^3 + 1080 x^2 + 810 x + 243$

$(2x-3y)^4=16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4$

The Pascal’s triangle is given below. In Pascal’s triangle nth row represents the coefficients of the expression of $(a+n)^{n-1}$.

In the expression $(x + 2)^6$, a=x, b=2 and n=6. Using 7th row of Pascal’s triangle we get

$(x + 2)^6=1(x^6)(2^0)+6(x^5)(2^1)+15(x^4)(2^2)+20(x^3)(2^3)+15(x^2)(2^4)+6(x^1)(2^5)+1(x^0)(2^6)$

$(x + 2)^6=x^6 + 12 x^5 + 60 x^4 + 160 x^3 + 240 x^2 + 192 x + 64$

Similarly,

In $(x-4)^4$ a=x, b=-4 and n=4. Using 5th row of Pascal’s triangle we get

$(x-4)^4=x^4 - 16 x^3 + 96 x^2 - 256 x + 256$

In $(2x+3)^5$ a=2x, b=3 and n=5. Using 6th row of Pascal’s triangle we get

$(2x+3)^5=32 x^5 + 240 x^4 + 720 x^3 + 1080 x^2 + 810 x + 243$

In $(2x-3y)^4$ a=2x, b=-3y and n=4. Using 5th row of Pascal’s triangle we get

$(2x-3y)^4=16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4$

In binomial expansion of $(3a + 4b)^8$ the degree of each term must be 8. It means the sum of powers of a and b is equal to 8.

For $a^2b^3$ the dgree of term is 2+3=5. So, it is not possible variable term.

Therefore the possible terms are $a^5b^3,b^8,a^4b^4,a^8,ab^7$.