Newton's binomial and Pascal's triangle

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

  1. $$\begin{array}{rl} (2a-b)^3=& \begin{pmatrix} 3 \\ 0 \end{pmatrix} (2a)^3 - \begin{pmatrix} 3 \\ 1 \end{pmatrix} (2a)^3 b + \begin{pmatrix} 3 \\ 2 \end{pmatrix} (2a)^2 b^2 - \begin{pmatrix} 3 \\ 3 \end{pmatrix} (2a) b^3 \\ =& 8a^3-12a^2b+6ab^2-b^3 \end{array}$$

  2. We will obtain the sixth term doing $k = 5$ in the formula: $$\begin{pmatrix} 10 \\ 5 \end{pmatrix} x^5(2y)^5= \dfrac{10!}{5!5!}x^5 32 y^5=8064 x^5 y^5$$

  3. 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: $$\begin{pmatrix} 8 \\ 4 \end{pmatrix} (3x^2)^4(ay)^4= \dfrac{8!}{4!4!}81 x^8 a^4 y^4=5670 x^8 a^4 y^4$$

  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.

  1. $ 8a^3-12a^2b+6ab^2-b^3 $
  2. $8064 x^5 y^5$
  3. $5670 x^8 a^4 y^4$
  4. 7th term.
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