If - - are roots of 375 x 2-25 x -2-0 and S n - n - n- then the value of 36 lim n - r -1 n S r is

375x^2 25x 2=0 α+β=115, αβ= 2375 lim_n→∞∑_r=1^nS_r= lim_n→∞∑_r=1^n(α^r+β^r) nbsp; =(α+α^2+α^3+...)+(β+β^2+β^3+...) =α1 α+β1 β =α+β 2αβ1 (α+β)+αβ =115+43751 ...

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Standard XII

Mathematics

Determinant

If α, β are r...

Question

If $α, β$ are roots of $375 x^{2} - 25 x - 2 = 0$ and $S_{n} = α^{n} + β^{n}$ , then the value of $36 lim n \to \infty n \sum r = 1 S_{r}$ is

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Solution

$375 x^{2} - 25 x - 2 = 0$
$α + β = \frac{1}{15}, α β = - \frac{2}{375}$

$lim n \to \infty n \sum r = 1 S_{r} = lim n \to \infty n \sum r = 1 (α^{r} + β^{r})$
$= (α + α^{2} + α^{3} + . . .) + (β + β^{2} + β^{3} + . . .)$
$= \frac{α}{1 - α} + \frac{β}{1 - β}$
$= \frac{α + β - 2 α β}{1 - (α + β) + α β}$
$= \frac{\frac{1}{15} + \frac{4}{375}}{1 - \frac{1}{15} - \frac{2}{375}}$
$= \frac{1}{12}$

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If α,β are roots of 375x2−25x−2=0 and Sn=αn+βn, then the value of 36limn→∞n∑r=1Sr is

375x2−25x−2=0 α+β=115, αβ=−2375 limn→∞n∑r=1Sr=limn→∞n∑r=1(αr+βr) =(α+α2+α3+...)+(β+β2+β3+...) =α1−α+β1−β =α+β−2αβ1−(α+β)+αβ =115+43751−115−2375 =112

If $α, β$ are roots of $375 x^{2} - 25 x - 2 = 0$ and $S_{n} = α^{n} + β^{n}$ , then the value of $36 lim n \to \infty n \sum r = 1 S_{r}$ is