Find numerically the greatest term in the expansion of (3−5x)15 when x=15
A
445×312
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B
445×310
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C
455×310
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D
455×312
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Solution
The correct option is D455×312 (3−5x)15 =315(1−5x3)15 Now At x=15 we get 315(1−13)15 Hence numerically greatest term will be given by r=(1+n)|x|1+|x| 16.1343 =4 Hence, T5=315.15C4(13)4 =311.1365 =455×312.