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
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