Question 1

I really need help solving thisIt’s from my trig prep bookIt asks to answer (a) and (b)

Question 2

Answer:

The sum in summation notation is:

$\sum ^4_{r\mathop{=}0}(-1)^r(4Crx^(4-r)y^r)$

The expansion is:

81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12)

Step-by-step explanation:

Given the expression:

In summation notation, this can be written as:

$\sum ^4_(r\mathop=0)(-1)^r(4Crx^(4-r)y^r)$

The simplified terms of the expression is:

$\begin{gathered} 4C0(3x^5)^{\mleft\{4-0\mright\}}((1)/(9)y^3)^0-4C1(3x^5)^{\mleft\{4-1\mright\}}((1)/(9)y^3)^1+4C2(3x^5)^{\mleft\{4-2\mright\}}((1)/(9)y^3)^2-4C3(3x^5)^{\mleft\{4-3\mright\}}((1)/(9)y^3)^3+4C4(3x^5)^{\mleft\{4-4\mright\}}((1)/(9)y^3)^4 \\ \\ =(3x^5)^4-4(3x^5)^3((1)/(9)y^3)+6(3x^5)^2((1)/(9)y^3)^2-4(3x^5)^{}((1)/(9)y^3)^3+((1)/(9)y^3)^4 \\ \\ =81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12) \end{gathered}$

0 Comments

Your comment on this question:

Your answer

1 Answer

0 Comments

Your comment on this answer:

Other Questions