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Q.

Numerically greatest term in the expansion of (3+2x)14 when x=15 is

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a

T1

b

T2

c

T3

d

T4

answer is B.

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Detailed Solution

(3+2x)14=3141+2x314 Here n=14;X=2x3=23×15=215Now(n+1)XX+1=(14+1)×215215+1 =21715 =3017 =1.7 =1+0.7 =P+F∴ Numerically greatest term in TP+1=T1+1=T2

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