The question is a bit ambiguous: you may mean
or
Anyway, in both cases, the domain of a function involving a square root is found by imposing the content of the root to be greateer than or equal to zero. So, in the first case you have
In the second case, you have
9514 1404 393
Answer:
64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729
Step-by-step explanation:
The row of Pascal's triangle we need for a 6th power expansion is ...
1, 6, 15, 20, 15, 6, 1
These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.
Your expansion is ...
1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...
15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6
= 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729
