Respondido

utilizado a teorema do binômio de newton desenvolver (2×+1)5​

Resposta :

EINSTEINDOYAHOOGO

(2x+1)⁵

=C5,0 *(2x)⁵⁻⁰ *1⁰ +C5,1 *(2x)⁵⁻¹ *1¹ +C5,2 *(2x)⁵⁻² *1² +C5,3 *(2x)⁵⁻³ *1³ +C5,4 *(2x)⁵⁻⁴ *1⁴ +C5,5 *(2x)⁵⁻⁵ *1⁵

=1*(2x)⁵ *1⁰ +5 *(2x)⁴ *1¹ +10 *(2x)³ *1² +10 *(2x)² *1³ +5 *(2x)¹ *1⁴ +1 *(2x)⁰*1⁵

= (2x)⁵ +5*(2x)⁴ + 10*(2x)³ + 10*(2x)²+5*(2x) + 1

= 32x⁵ +80x⁴+80x³+40x²+10x+1

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