If - are roots of 375x2-25x-2-0 and s- then lim-r-1- sr is

The correct option is B 1/12 α amp; β are the roots of the equation 375x^ 2 25x 2=0 Therefore, α +β = 25 375 nbsp; amp; ...

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Standard XII

Mathematics

Roots of a Quadratic Equation

If α,β are ...

Question

If $α, β$ are roots of $375 x^{2} - 25 x - 2 = 0$ and $s_{η} = α^{η} + β^{η}$ , then ${lim}_{η \to \infty} \sum_{r = 1}^{η} s_{r}$ is

\frac{7}{116}

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\frac{1}{12}

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\frac{29}{358}

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none of these

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Solution

The correct option is B $\frac{1}{12}$
$α$ & $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$
Therefore, $α + β = \frac{25}{375}$ & $α β = \frac{- 2}{375}$
Now, $S = lim η \to \infty η \sum r = 1 s_{r} = \infty \sum r = 1 (α^{r} + β^{r}) = \frac{α}{1 - α} + \frac{β}{1 - β} = \frac{α + β - 2 α β}{1 - α - β + α β}$
$\Rightarrow S = \frac{(25 / 375) - 2 (- 2 / 375)}{1 - (25 / 375) - (2 / 375)} = \frac{1}{12}$
Ans: B

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If α,β are roots of 375x2−25x−2=0 and sη=αη+βη, then limη→∞∑ηr=1sr is

The correct option is B 112α & β are the roots of the equation 375x2−25x−2=0Therefore, α+β=25375 & αβ=−2375Now, S=limη→∞η∑r=1sr=∞∑r=1(αr+βr)=α1−α+β1−β=α+β−2αβ1−α−β+αβ⇒S=(25/375)−2(−2/375)1−(25/375)−(2/375)=112Ans: B

If $α, β$ are roots of $375 x^{2} - 25 x - 2 = 0$ and $s_{η} = α^{η} + β^{η}$ , then ${lim}_{η \to \infty} \sum_{r = 1}^{η} s_{r}$ is