Question

Resolver por el binomio de Newton a. (2x+2)^5 b. (3x + 1)^6

Answer 1

Respuesta:

a)

$(2x+2)^5= 2^5(x+1)^5 = 32\left[\binom{5}{0}\cdot x^5\cdot 1^0 + \binom{5}{1}\cdot x^4\cdot 1^1 + \binom{5}{2}\cdot x^3\cdot 1^2 +$

$+ \left \binom{5}{3}\cdot x^2\cdot 1^3 + \binom{5}{4}\cdot x^1\cdot 1^4 + \binom{5}{5}\cdot x^0\cdot 1^5 \right]$

$=32(x^5+5x^4+10x^3+10x^2+5x+1)$

$=32x^5+160x^4+320x^3+320x^2+160x+32$

b)

$(3x+1)^6=\binom{6}{0}\cdot (3x)^6\cdot 1^0 + \binom{6}{1}\cdot (3x)^5\cdot 1^1 + \binom{6}{2}\cdot (3x)^4\cdot 1^2 +$

$+ \binom{6}{3}\cdot (3x)^3\cdot 1^3 + \binom{6}{4}\cdot (3x)^2\cdot 1^4 + \binom{6}{5}\cdot (3x)^1\cdot 1^5 + \binom{6}{6}\cdot (3x)^0\cdot 1^6$

$=729x^6+1458x^5+1215x^4+540x^3+135x^2+18x+1$