Problems from Newton's binomial and Pascal's triangle

Calculate the development of $(2 a - b)^{3}$ .
Calculate the sixth term of $(x + 2 y)^{10}$ .
Find the central term of $(3 x^{2} + a y)^{8}$ .
What is the term that contains $x^{20}$ in the development of $(x^{2} - x y)^{13}$ ?

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$\begin{aligned} (2 a - b)^{3} = & (\begin{array}{c} 3 \\ 0 \end{array}) (2 a)^{3} - (\begin{array}{c} 3 \\ 1 \end{array}) (2 a)^{3} b + (\begin{array}{c} 3 \\ 2 \end{array}) (2 a)^{2} b^{2} - (\begin{array}{c} 3 \\ 3 \end{array}) (2 a) b^{3} \\ = & 8 a^{3} - 12 a^{2} b + 6 a b^{2} - b^{3} \end{aligned}$
We will obtain the sixth term doing $k = 5$ in the formula: $(\begin{matrix} 10 \\ 5 \end{matrix}) x^{5} (2 y)^{5} = \frac{10!}{5! 5!} x^{5} 32 y^{5} = 8064 x^{5} y^{5}$
The development has $9$ terms, then the central is the one that will occupy the fifth place, or the one that is obtained by doing $k = 4$ in the general formula: $(\begin{matrix} 8 \\ 4 \end{matrix}) x^{4} (2 y)^{4} = \frac{8!}{4! 4!} 81 x^{8} a^{4} y^{4} = 5670 x^{8} a^{4} y^{4}$
We have to calculate the value of $k$ : $(x^{2})^{13 - k} x^{k} = x^{2 (13 - k)} x^{k} = x^{26 - k}$ $x^{26 - k} = x^{20} \Rightarrow 26 - k = 20 \Rightarrow k = 6$ It will be, therefore, the seventh term.

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