If a1,a2,a3....an∈R+ and a1.a2.a3....an=1, then minimum value of (1+a1+a21)(1+a2+a22)(1+a3+a23)....(1+an+a2n) is equal to.
A
3n+1
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B
3n
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C
3n−1
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D
none of these
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Solution
The correct option is B3n The given expression will be minimum, when a1=a2=a3=...an. Now, it is given that a1.a2.a3...an=1 Hence, ani=1 Since ai is Real, therefore ai=1 Hence, (1+a1+a21)(1+a2+a22)...(1+an+a2n) =(1+ai+a2i)n =3n.