Middle Term Binomial 683299

1. **State the problem:** Find the middle term of the binomial expansion of $$ (2x + y)^6 $$ and then evaluate it at $$ x=3 $$ and $$ y=\frac{1}{2} $$. 2. **Recall the binomial expansion formula:** $$ (a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r $$ The middle term for an expansion with even $$ n $$ is the $$ \left( \frac{n}{2} + 1 \right)^{th} $$ term. 3. **Identify the middle term:** Since $$ n=6 $$, the middle term is the $$ \left( \frac{6}{2} + 1 \right) = 4^{th} $$ term. 4. **Write the general term:** $$ T_{r+1} = \binom{6}{r} (2x)^{6-r} y^r $$ For the 4th term, $$ r=3 $$: $$ T_4 = \binom{6}{3} (2x)^{3} y^{3} $$ 5. **Calculate the binomial coefficient:** $$ \binom{6}{3} = \frac{6!}{3!3!} = 20 $$ 6. **Substitute and simplify:** $$ T_4 = 20 \times (2x)^3 \times y^3 = 20 \times 8x^3 \times y^3 = 160 x^3 y^3 $$ 7. **Evaluate at $$ x=3 $$ and $$ y=\frac{1}{2} $$:** $$ T_4 = 160 \times 3^3 \times \left( \frac{1}{2} \right)^3 = 160 \times 27 \times \frac{1}{8} $$ 8. **Simplify:** $$ 160 \times 27 \times \frac{1}{8} = \cancel{160}^{20} \times 27 \times \frac{1}{\cancel{8}} = 20 \times 27 = 540 $$ **Final answer:** The middle term is $$ 160 x^3 y^3 $$ and its value at $$ x=3 $$ and $$ y=\frac{1}{2} $$ is $$ 540 $$.