A) \[55\times {{3}^{9}}\]
B) \[55\times {{3}^{6}}\]
C) \[45\times {{3}^{6}}\]
D) \[45\times {{3}^{9}}\]
Correct Answer: A
Solution :
We have, \[{{(3-5x)}^{11}}={{3}^{11}}{{\left( 1-\frac{5x}{3} \right)}^{11}}\] \[={{3}^{11}}{{\left( 1-\frac{5}{3}.\frac{1}{5} \right)}^{11}}\] \[\left\{ \because \,x=\frac{1}{5} \right\}\] \[={{3}^{11}}{{\left( 1-\frac{1}{3} \right)}^{11}}\] Now, \[r=\frac{\left| \,x\, \right|\,(n+1)}{\left| \,x\, \right|+1}\] \[=\frac{\left| -\frac{1}{3} \right|\,(11+1)}{\left| -\frac{1}{3} \right|+1}\] \[=\frac{4}{4/3}\] \[\Rightarrow \] \[r=3\] Therefore, \[3rd({{T}_{3}})\]and \[(3+1)=4th({{T}_{4}})\]terms are numerically greatest in the expansion of \[{{(3-5x)}^{11}}.\] Hence, greatest term\[={{T}_{3}}\] \[={{3}^{11}}\left| {}^{11}{{C}_{2}}{{(1)}^{9}}{{\left( -\frac{1}{3} \right)}^{2}} \right|\] \[={{3}^{11}}\left| \frac{11\times 10}{1\cdot 2\cdot 9} \right|\] \[=55\times {{3}^{9}}\] and \[{{T}_{4}}={{3}^{11}}\left| {}^{11}{{C}_{3}}{{(1)}^{8}}{{\left( -\frac{1}{3} \right)}^{3}} \right|\] \[={{3}^{11}}\left| \frac{11\times 10\times 9}{1\cdot 2\cdot 3}\cdot \left( -\frac{1}{27} \right) \right|\] \[=55\times {{3}^{9}}\] Hence, greatest term (numerically)\[={{T}_{3}}={{T}_{4}}\] \[=55\times {{3}^{9}}\]
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