Answer :

konrad509
[tex]T_r={n \choose r-1}a^{n-(r-1)}b^{r-1}\\ T_6={9 \choose 6-1}(2x)^{9-(6-1)}y^{6-1}\\ T_6={9 \choose 5}(2x)^4y^{5}\\ T_6=\frac{9!}{5!4!}16x^4y^{5}\\ T_6=\frac{6\cdot7\cdot8\cdot9}{2\cdot3\cdot4}16x^4y^{5}\\ T_6=7\cdot2\cdot9\cdot16x^4y^{5}\\ T_6=2016x^4y^5[/tex]
[tex]Use\ the\ Pascal's\ Triangle\ (look\ at\ the\ picture)\\\\6-th\ term\ of\ (2x+y)^9\ is\\\\ 126(2x)^4y^5=126\times16x^4y^5=2016x^4y^5[/tex]
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