Question

Find the middle term in the expansion of $[\frac{x}{3}+9y]^{10}$

• A. $$=41003x^6y^4$$
• B. $$=51030x^4y^6$$
• C. $$=60480x^5y^5$$
• D. $$=61236x^5y^5$$

Answer 1

Answer:

$$=61236x^5y^5$$

$[\frac{x}{3}+9y]^{10}$

\frac{n}{2}+1)^{th}=>(\frac{10}{2}+1)^{th}=>6^{th}

$a=x/3, b=9y, n=10, r+1=6, r=5, n-r=5$

$T_{r+1}=^nC_ra^{n-r}b^r$

$T_6=^{10}C_5(\frac{x}{3})^5(9y)^5$

$T_6=\frac{10!}{5! \times 5!}\times\frac{x^5}{3^5}\times9^5y^5$

$=\frac{10\times9 \times 8\times7\times6\times5!}{5\times4\times3\times2\times1 \times5!}\times \frac{9\times9\times9\times9\times9}{3\times3\times3\times3\times3} \times x^5y^5$

$=3\times2\times7\times6\times3^5x^5y^5$

$=252 \times 243 x^5y^5$

$=61236x^5y^5$