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Question

If α,β are the roots of the equation ax2+bx+c=0,a0 and Sn=αn+βn, then

A
aSn+1+bSn+cSn1=0
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B
S5=ba5(b22ac)2(b2ac)bca4
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C
aSn+1+bSn+cSn1=1
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D
S5=ba5(b22ac)2+(b2ac)bca4
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Solution

The correct option is D S5=ba5(b22ac)2+(b2ac)bca4
We know that,
α,β are the roots of the equation ax2+bx+c=0,a0,
aα2+bα+c=0
aβ2+bβ+c=0
α+β=ba, αβ=ca

Now,
Sn=αn+βn
aSn+1+bSn+cSn1
=a(αn+1+βn+1)+b(αn+βn)+c(αn1+βn1)
=αn1(aα2+bα+c)+βn1(aβ2+bβ+c)
=αn1×0+βn1×0
=0

Sn+1=baSncaSn1 ...(1)
Putting n=4 in the above equation, we get
S5=baS4caS3
Now,
S4=α4+β4 =((α+β)22αβ)22α2β2 =(b2a22ca)22c2a2 =(b22ac)2a42c2a2
S3=α3+β3 =(α+β)33αβ(α+β) =b3a3+3bca2
Putting in the expression of S5,
S5=ba5(b22ac)2+2bc2a3+b3ca43bc2a3S5=ba5(b22ac)2+b3ca4bc2a3

S5=ba5(b22ac)2+(b2ac)bca4


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