The correct answers are:
1) d⁶-18d⁵+135d⁴-540d³+1215d²-1458d+729
2) 243v⁵+405v⁴s+270v³s²+90v²s³+15vs⁴+s⁵
3) -64b³
Explanation:
We first write out Pascal's Triangle. We start with 1; below this is a row of two 1's; from this line down, we start with 1 and each successive term is the sum of the two above it. I have attached a picture of Pascal's Triangle.
The coefficients of our expanded polynomial are the numbers in each row of the triangle. The first row represents the coefficients of a binomial to the 0 power; the second row represents the 1st power; the third row, the 2nd power; etc.
Using this, the coefficients of a binomial to the 6th power are: 1, 6, 15, 20, 15, 6, 1.
As we expand our polynomial, our first term will be the first coefficient multiplied by the first term of the binomial to the 6th power and the second term to the 0 power. The second term will be the second coefficient multiplied by the first term of the binomial to the 5th power and the second term to the 1st power; the third term will be the third coefficient multiplied by the first term of the binomial to the 4th power and the second term to the 2nd power; etc. In each term of the expansion, the first term of the binomial loses a power while the second term gains a power. This gives us:
1d⁶(-3)⁰+6(d⁵(-3)¹)+15(d⁴(-3)²)+20(d³(-3)³)+15(d²(-3)⁴)+6(d¹(-3)⁵)+1(d⁰(-3)⁶)
= d⁶-18d⁵+15(9d⁴)+20(-27d³)+15(81d²)+6(-243d)+729
= d⁶-18d⁵+135d⁴-540d³+1215d²-1458d+729
2) The coefficients for the 5th power, based on Pascal's Triangle, would be:
1, 5, 10, 10, 5, 1
This gives us:
1(3v)⁵s⁰+5(3v)⁴s¹+10(3v)³s²+10(3v)²s³+5(3v)¹s⁴+1(3v)⁰s⁵
= 243v⁵+5(81v⁴s)+10(27v³s²)+10(9v²s³)+5(3vs⁴)+s⁵
= 243v⁵+405v⁴s+270v³s²+90v²s³+15vs⁴+s⁵
3) The coefficients of the third power in Pascal's Triangle would be:
1, 3, 3, 1
Since there are 4 terms in this row, the first term will have d³ and (-4b)⁰; in each successive term, d will lose an exponent and (-4b) will gain an exponent. In the fourth term, d will have an exponent of 0 and (-4b) will have an exponent of 3; this makes the fourth term
(-4b)³ = -64b³
Answer:
1) option:A
2) [tex](3v+s)^5=243v^5+405v^4s+270v^3s^2+90v^2s^3+15vs^4+s^5[/tex]
3) [tex]-64b^3[/tex]
Step-by-step explanation:
The binomial expansion of [tex](ax+by)^n[/tex] is given by:
[tex](ax+by)^n=\binom{n}{0} ax^{n-1} (by)^0+\binom{n}{1} (ax)^{n-1}(by)^{1}+\binom{n}{1}(ax)^{n-2}(by)^{2}+.......+\binom{n}{n}(ax)^{0}(by)^{n}[/tex]
1) on finding the expansion of [tex](d-3)^6[/tex].
[tex]((d-3)^6)=(d^3-27-9d^2+27d)^2=d^6-18d^5+135d^4-540d^3+1215d^2-1458d+729[/tex]
Hence option A is correct.
2) on finding the expansion of
[tex](3v+s)^5[/tex]
Hence,
[tex](3v+s)^5=243v^5+405v^4s+270v^3s^2+90v^2s^3+15vs^4+s^5[/tex]
3) We need to find the fourth term of [tex](d-4b)^3[/tex] i.e.the term of [tex]b^3[/tex].
[tex](d-4b)^3=d^3+(-4b)^3+3d^2\times(-4b)+3d\times\\(-4b)^2\\(d-4b)^3=d^3-64b^3-12d^2b+48db^2[/tex]
hence, the fourth term is: [tex]-64b^3[/tex]
