Answer:
a) [tex](3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5[/tex].
b) The middle term in the expansion is [tex]\frac{6}{x}[/tex].
c) The coefficient of [tex]x^{11}[/tex] is 120.
Step-by-step explanation:
Remember that the binomial theorem say that [tex](x+y)^n=\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^{k}[/tex]
a) [tex](3x+y)^5=\sum_{k=0}^5\binom{5}{k}3^{n-k}x^{n-k}y^k[/tex]
Expanding we have that
[tex]\binom{5}{0}3^5x^5+\binom{5}{1}3^4x^4y+\binom{5}{2}3^3x^3y^2+\binom{5}{3}3^2x^2y^3+\binom{5}{4}3xy^4+\binom{5}{5}y^5[/tex]
symplifying,
[tex](3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5[/tex].
b) The middle term in the expansion of [tex](\frac{1}{x} +\sqrt{x})^4=\sum_{k=0}^{4}\binom{4}{k}\frac{1}{x^{4-k}}x^{\frac{k}{2}}[/tex] correspond to k=2. Then [tex]\binom{4}{2}\frac{1}{x^2}x^{\frac{2}{2}}=\frac{6}{x}[/tex].
c) [tex](x^2+\frac{1}{x})^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{2(10-k)}\frac{1}{x^k}=\sum_{k=0}^{10}\binom{10}{k}x^{20-2k}\frac{1}{x^k}=\sum_{k=0}^{10}\binom{10}{k}x^{20-3k}[/tex]
Since we need that 11=20-3k, then k=3.
Then the coefficient of [tex]x^{11}[/tex] is [tex]\binom{10}{3}=120[/tex]
