Harold uses the binomial theorem to expand the binomial [tex](3x^5 -\dfrac{1}{9}y^3)^4[/tex]
(a) What is the sum in summation notation that he uses to express the expansion?
(b) Write the simplified terms of the expansion.
Answer:
(a). [tex](3x^5 -\dfrac{1}{9}y^3)^4=$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $[/tex]
(b).[tex](3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}[/tex]
Step-by-step explanation:
(a).
The binomial theorem says
[tex](x+y)^n=$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $[/tex]
For our binomial this gives
[tex]\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $}[/tex]
(b).
We simplify the terms of the expansion and get:
[tex]$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4[/tex]
[tex]$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4[/tex]
[tex]\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561} }[/tex]
