Question

Utilizando o Teorema do Binômio de Newton, desenvolver.
a) (x + y) elevado a 3
b) (2x + 1) elevado a 6

(me ajudem pfv)

Answer 1

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Binômio de Newton

$\displaystyle\sf (a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+...\binom{n}{n}b^n$

$\displaystyle\tt a)~\sf (x+y)^3=\binom{3}{0}x^3+\binom{3}{1}x^2y+\binom{3}{2}xy^2+\binom{3}{3}y^3\\\sf\binom{3}{0}=1\\\sf\binom{3}{1}\dfrac{3}{1!}=3\\\sf\binom{3}{2}=\dfrac{3\cdot2}{2!}=3\\\sf\binom{3}{3}=1\\\sf (x+y)^3=x^3+3x^2y+3xy^2+y^3$

$\displaystyle\tt b)~\sf (2x+1)^6=\binom{6}{0}(2x)^6+\binom{6}{1}(2x)^5\cdot1+\binom{6}{2}(2x)^4\cdot1^2+\binom{6}{3}(2x)^3\cdot 1^3+\binom{6}{4}(2x)^2\cdot1^4+\binom{6}{5}(2x)\cdot1^5+\binom{6}{6}1^6\\\underline{\tt c\acute alculos~auxiliares:}\\\sf\binom{6}{0}=1\\\sf\binom{6}{1}=6\\\sf\binom{6}{2}=\dfrac{6\cdot5}{2\cdot1}=15\\\sf\binom{6}{3}=\dfrac{6\cdot5\cdot4}{3\cdot2\cdot1}=20\\\sf\binom{6}{4}=\dfrac{6\cdot5\cdot4\cdot3}{4\cdot3\cdot2\cdot1}=15$

$\displaystyle\sf\binom{6}{5}=\dfrac{6\cdot5\cdot4\cdot3\cdot2}{5\cdot4\cdot3\cdot2}=6\\\sf\binom{6}{6}=1$

$\tt b)~\sf(2x+1)^6=64x^6+6\cdot32x^5+15\cdot16x^4+20\cdot8x^3+15\cdot4x^2+6\cdot2x+1\\\sf(2x+1)^6=64x^6+192x^5+240x^4+160x^3+60x^2+12x+1$