If α, β are the roots of the equation 375 x2-25 x-2=0 and Sn=αn+βn, then undersetn arrow ∞ textLt displaystyle∑r=1n Sr is

Question

If α, β are the roots of the equation 375 x2-25 x-2=0 and Sn=αn+βn, then undersetn arrow ∞ textLt displaystyle∑r=1n Sr is (A) (7/12) (B) (1/12) (C) (35/12) (D) None of these

Tardigrade Admin · Accepted Answer

displaystyle∑r=1n Sr=(α+β)+(α2+β2)+ ldots+(αn+βn) =(α+α2+ ldots+αn)+(β+β2+ ldots+βn) undersetn arrow ∞ textLt displaystyle∑r=1n Sr=(α+α2+ ldots+∞)+(β+β2+ ldots+∞) =(α/1-α)+(β/1-β) =(α-α β+β-α β/1-(α+β)+α β) =(α+β-2 α β/1-(α+β)+α β) =((25/375)+(4/375)/1-(25)375-(2)375 =(29/348)=(1/12)

Q. If $α, β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ and $S_{n} = α^{n} + β^{n}$ , then $n \to \infty Lt r = 1 \sum n S_{r}$ is

$\frac{7}{12}$

$\frac{1}{12}$

$\frac{35}{12}$

None of these

Q. If α,β are the roots of the equation 375x2−25x−2=0 and Sn​=αn+βn, then n→∞Lt​r=1∑n​Sr​ is

127​

121​

1235​

None of these

Q. If $α, β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ and $S_{n} = α^{n} + β^{n}$ , then $n \to \infty Lt r = 1 \sum n S_{r}$ is

$\frac{7}{12}$

$\frac{1}{12}$

$\frac{35}{12}$