Numerically greatest term in (1−2y)5 when y=12

( 1 − 2 y ) 5 m = | 2 y | ( n + 1 ) | 2 y | + 1 = 2 2 × 6 2 2 + 1 = 6 2 = 3 T 3 , T 3 + 1 3 rd term, 4 th term

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Numerically greatest term in ${(1 - 2 y)}^{5} when y = \frac{1}{2}$

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3^rd

4^th

5^th

2^nd

answer is B, C.

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${(1 - 2 y)}^{5}$

$m = \frac{| 2 y | (n + 1)}{| 2 y | + 1}$

$= \frac{\frac{2}{2} \times 6}{\frac{2}{2} + 1} = \frac{6}{2} = 3$

$T_{3}, T_{3 + 1}$

3^rd term, 4^thterm

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