Greatest term!

${ (2+3x) }^{ 9 }={ 2 }^{ 9 }{ (1+\frac { 3x }{ 2 } ) }^{ 9 },\\ To\quad find\quad greatest\quad term\quad of\quad expansion\quad means\\ to\quad find\quad greatest\quad value\quad of\quad 'r'\quad for\quad which\\ { T }_{ r+1 }\ge { T }_{ r }.\\ For\quad given\quad expansion,\\ \frac { { T }_{ r+1 } }{ { T }_{ r } } =\frac { (9+1-r) }{ r } \frac { 3x }{ 2 } =\frac { (9+1-r) }{ r } \frac { 3 }{ 2 } \frac { 3 }{ 2 } \ge 1\\ \therefore 90\quad -\quad 9r\quad \ge \quad 4r\quad \therefore \quad r\quad \le \quad 6.92\\ but\quad r\in N,\quad \therefore \quad r\quad =\quad 6.\\ \therefore \quad { T }_{ r+1 }\quad =\quad { T }_{ 7 }\quad is\quad the\quad gretest\quad term.\\ { T }_{ 7 }\quad =\quad { c }_{ 6 }^{ 9 }\quad { 1 }^{ 3 }{ (\frac { 9 }{ 4 } ) }^{ 6 }{ 2 }^{ 9 }=\quad \frac { 9\times 8\times 7 }{ 3\times 2\times 1 } \frac { { 3 }^{ 12 } }{ { 2 }^{ 12-9 } } =\frac { { 3 }^{ 13 }\times 7 }{ 2 } \\ \therefore \quad a\quad =\quad 13,\quad b\quad =\quad 7,\quad c\quad =\quad 2\\ \therefore a+b+c\quad =\quad 22$