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Question

If α, β are roots of ax2+bx+c=0 and D=b24ac, and Sn=1+αn+βn then
∣ ∣S0S1S2S1S2S3S2S3S4∣ ∣

A
1
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B
D(a+b+c)3a4
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C
(b24ac)2a4
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D
D(a+b+c)2a4
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Solution

The correct option is C D(a+b+c)2a4
∣ ∣S0S1S2S1S2S3S2S3S4∣ ∣=∣ ∣ ∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣ ∣ ∣=∣ ∣1111αβ1α2β2∣ ∣∣ ∣1111αβ1α2β2∣ ∣=∣ ∣1111αβ1α2β2∣ ∣2=(1α)2(1β)2(αβ)2=(1αβ+αβ)2((α+β)24αβ)=(1(ba)+ca)2((ba)24ca)=D(a+b+c)2a4

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