Desenvolva, usando o teorema binomial:
\left(1+\frac{3 a}{2}\right)^{4}

Passo 1

O enunciado pede para expandir o binômio \displaystyle\bigl(1+\tfrac{3a}{2}\bigr)^{4} usando o Teorema Binomial. Vamos aplicar a fórmula geral de expansão de \,(x+y)^{n}.

Passo 2

O Teorema Binomial diz que, para inteiro n\ge0,
(x+y)^{n} \;=\;\sum_{k=0}^{n} \binom{n}{k}\,x^{\,n-k}\,y^{\,k}\,,
onde \displaystyle\binom{n}{k}=\frac{n!}{k!\,(n-k)!}.

Passo 3

Aqui temos n=4, x=1 e y=\tfrac{3a}{2}. Logo,
\bigl(1+\tfrac{3a}{2}\bigr)^{4} =\sum_{k=0}^{4}\binom{4}{k}\,1^{\,4-k}\,\Bigl(\tfrac{3a}{2}\Bigr)^{k} =\sum_{k=0}^{4}\binom{4}{k}\,\Bigl(\tfrac{3a}{2}\Bigr)^{k}\,.

Passo 4

Agora calculamos cada termo k=0,1,2,3,4:
\begin{aligned} k=0:&\quad \binom{4}{0}\Bigl(\tfrac{3a}{2}\Bigr)^{0}=1,\\\ k=1:&\quad \binom{4}{1}\Bigl(\tfrac{3a}{2}\Bigr)^{1} =4\cdot\tfrac{3a}{2}=6a,\\\ k=2:&\quad \binom{4}{2}\Bigl(\tfrac{3a}{2}\Bigr)^{2} =6\cdot\tfrac{9a^{2}}{4}=\tfrac{54a^{2}}{4}=\tfrac{27a^{2}}{2},\\\ k=3:&\quad \binom{4}{3}\Bigl(\tfrac{3a}{2}\Bigr)^{3} =4\cdot\tfrac{27a^{3}}{8}=\tfrac{108a^{3}}{8}=\tfrac{27a^{3}}{2},\\\ k=4:&\quad \binom{4}{4}\Bigl(\tfrac{3a}{2}\Bigr)^{4} =1\cdot\tfrac{81a^{4}}{16}=\tfrac{81a^{4}}{16}. \end{aligned}
Somando os termos, obtemos a expansão completa.

Resposta

\bigl(1+\tfrac{3a}{2}\bigr)^{4} =1 \;+\;6a \;+\;\tfrac{27a^{2}}{2}\;+\;\tfrac{27a^{3}}{2}\;+\;\tfrac{81a^{4}}{16}.

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