- Calculate the development of $$(2a-b)^3$$.
- Calculate the sixth term of $$(x+2y)^{10}$$.
- Find the central term of $$(3x^2+ay)^8$$.
- What is the term that contains $$x^{20}$$ in the development of $$(x^2-xy)^{13}$$?
Development:
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$$$\begin{array}{rl} (2a-b)^3=& \begin{pmatrix} 3 \\ 0 \end{pmatrix} (2a)^3 - \begin{pmatrix} 3 \\ 1 \end{pmatrix} (2a)^3 b + \begin{pmatrix} 3 \\ 2 \end{pmatrix} (2a)^2 b^2 - \begin{pmatrix} 3 \\ 3 \end{pmatrix} (2a) b^3 \\ =& 8a^3-12a^2b+6ab^2-b^3 \end{array}$$$
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We will obtain the sixth term doing $$k = 5$$ in the formula: $$$\begin{pmatrix} 10 \\ 5 \end{pmatrix} x^5(2y)^5= \dfrac{10!}{5!5!}x^5 32 y^5=8064 x^5 y^5$$$
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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: $$$\begin{pmatrix} 8 \\ 4 \end{pmatrix} x^4(2y)^4= \dfrac{8!}{4!4!}81 x^8 a^4 y^4=5670 x^8 a^4 y^4$$$
- 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.
Solution:
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$$ 8a^3-12a^2b+6ab^2-b^3 $$
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$$8064 x^5 y^5$$
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$$5670 x^8 a^4 y^4$$
- 7th term.