Saved Bookmarks
| 1. |
Find the term independent of x in the expansion (x/√3)+(√3/2x²)^10\xa0 |
| Answer» {tex}{\\left( {\\sqrt {{x \\over 3}} + {3 \\over {2{x^2}}}} \\right)^{10}}{/tex}{tex}{T_{r + 1}}{ = ^{10}}{C_r}{\\left( {\\sqrt {{x \\over 3}} } \\right)^{10 - r}}{\\left( {{3 \\over {2{x^2}}}} \\right)^r}{/tex}{tex}{ = ^{10}}{C_r}{\\left( x \\right)^{5 - {r \\over 2} - 2r}}{\\left( 3 \\right)^{ - 5 + {r \\over 2} + r}}{\\left( 2 \\right)^{ - r}}{/tex}{tex}{ = ^{10}}{C_r}{\\left( x \\right)^{{{10 - 5r} \\over 2}}}{\\left( 3 \\right)^{{{3r - 10} \\over 2}}}{\\left( 2 \\right)^{ - r}}{/tex}{tex}for\\,independent\\,of\\,x{/tex}{tex}{\\left( x \\right)^{{{10 - 5r} \\over 2}}} = {x^0}{/tex}{tex}{{10 - 5r} \\over 2} = 0{/tex}{tex}10 - 5r = 0{/tex}{tex}r = 2{/tex}{tex}independent\\,term\\,{T_3}{ = ^{10}}{C_2}{\\left( 3 \\right)^{ - 2}}{\\left( 2 \\right)^{ - 2}}{/tex}{tex} = {{10!} \\over {2!8!}} \\times {1 \\over {36}}{/tex}{tex} = {{10 \\times 9} \\over {2 \\times 36}}{/tex}{tex} = {5 \\over 4}{/tex} | |