Question

Find the constant term in the expansion of (2x^2 - 1/x)^6 (Binomial Expansion) Please show working out!

Answer 1

Recall that the T(k+1) term in the expansion of any binomial is given by

\[T_{k+1}= \left(\begin{matrix}n \\ k\end{matrix}\right) a^{n-k}b^k\]
For our case the power n = 6.

\[(2x^2 - \frac{ 1 }{ x })^6 = \sum_{k=0}^{6}(2x^2)^{6-k}(\frac{ -1 }{ x })^k\]
So each term is given by taking the powers and grouping any powers involving x. Remember your index laws for multiplying.
\[T_{k+1}=\left(\begin{matrix}6 \\ k\end{matrix}\right)(2x)^{6-k}(\frac{ -1 }{ x })^k = \left(\begin{matrix}6 \\ k\end{matrix}\right)2^{6-k}(-1)^k x^{6-k}x^{-k} =\left(\begin{matrix}6 \\ k\end{matrix}\right)2^{6-k}(-1)^kx^{6-2k}\]

Once you establish an equation for T(k+1) as I did above the next part is easy.
For a constant term, the power linked to x must be 0 since x^0 = 1 (a constant). So focus only on the term with x and not the other numbers and equate the power of x to 0. This will enable you to find k.

Putting this into the T(k+1) will give you the term independent of x.