if α and β are the roots of the equation 375x² – 25x -2 = 0 then [tex]\displaystyle \lim_{\sf n\to \infty }\sum^{n}_{r=1}\alpha^r + \lim_{\sf n\to \infty }\sum^{n}_{r=1}\beta^r[/tex] is equal to About the author Vivian
The roots of the equation [tex]375x^2-25x-2=0[/tex] are, [tex]\longrightarrow x=\dfrac{25\pm\sqrt{(-25)^2-(4\times375\times(-2))}}{2\times375}[/tex] [tex]\longrightarrow x=\dfrac{25\pm\sqrt{3625}}{750}[/tex] [tex]\longrightarrow x=\dfrac{5\pm\sqrt{145}}{150}[/tex] So, [tex]\alpha=\dfrac{5+\sqrt{145}}{150}[/tex] [tex]\beta=\dfrac{5-\sqrt{145}}{150}[/tex] We see [tex]|\alpha|<1,\ |\beta|<1[/tex] also. Then, [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\lim_{n\to\infty}\sum_{r=1}^n\left(\dfrac{5+\sqrt{145}}{150}\right)^r[/tex] The sum actually represents an infinite geometric series of first term [tex]\alpha[/tex] and common ratio also [tex]\alpha.[/tex] We have the formula for infinite geometric series, that for [tex]|k|<1,[/tex] [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^nak^{r-1}=\dfrac{a}{1-k}[/tex] Here [tex]a[/tex] is first term and [tex]k[/tex] is common ratio. If [tex]a=k[/tex] then, [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^nk^r=\dfrac{k}{1-k}[/tex] Thus, [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\dfrac{\frac{5+\sqrt{145}}{150}}{1-\frac{5+\sqrt{145}}{150}}[/tex] [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\dfrac{5+\sqrt{145}}{145-\sqrt{145}}[/tex] Similarly, [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\lim_{n\to\infty}\sum_{r=1}^n\left(\dfrac{5-\sqrt{145}}{150}\right)^r[/tex] [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{\frac{5-\sqrt{145}}{150}}{1-\frac{5-\sqrt{145}}{150}}[/tex] [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{5-\sqrt{145}}{145+\sqrt{145}}[/tex] Now, [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{5+\sqrt{145}}{145-\sqrt{145}}+\dfrac{5-\sqrt{145}}{145+\sqrt{145}}[/tex] [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{2(145\times5+145)}{145^2-145}[/tex] [tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{2\times145\times6}{145\times144}[/tex] [tex]\displaystyle\longrightarrow\underline{\underline{\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{1}{12}}}[/tex] Hence 1/12 is the answer. Reply
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1/12
The roots of the equation [tex]375x^2-25x-2=0[/tex] are,
[tex]\longrightarrow x=\dfrac{25\pm\sqrt{(-25)^2-(4\times375\times(-2))}}{2\times375}[/tex]
[tex]\longrightarrow x=\dfrac{25\pm\sqrt{3625}}{750}[/tex]
[tex]\longrightarrow x=\dfrac{5\pm\sqrt{145}}{150}[/tex]
So,
We see [tex]|\alpha|<1,\ |\beta|<1[/tex] also.
Then,
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\lim_{n\to\infty}\sum_{r=1}^n\left(\dfrac{5+\sqrt{145}}{150}\right)^r[/tex]
The sum actually represents an infinite geometric series of first term [tex]\alpha[/tex] and common ratio also [tex]\alpha.[/tex]
We have the formula for infinite geometric series, that for [tex]|k|<1,[/tex]
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^nak^{r-1}=\dfrac{a}{1-k}[/tex]
Here [tex]a[/tex] is first term and [tex]k[/tex] is common ratio.
If [tex]a=k[/tex] then,
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^nk^r=\dfrac{k}{1-k}[/tex]
Thus,
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\dfrac{\frac{5+\sqrt{145}}{150}}{1-\frac{5+\sqrt{145}}{150}}[/tex]
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r=\dfrac{5+\sqrt{145}}{145-\sqrt{145}}[/tex]
Similarly,
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\lim_{n\to\infty}\sum_{r=1}^n\left(\dfrac{5-\sqrt{145}}{150}\right)^r[/tex]
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{\frac{5-\sqrt{145}}{150}}{1-\frac{5-\sqrt{145}}{150}}[/tex]
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{5-\sqrt{145}}{145+\sqrt{145}}[/tex]
Now,
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{5+\sqrt{145}}{145-\sqrt{145}}+\dfrac{5-\sqrt{145}}{145+\sqrt{145}}[/tex]
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{2(145\times5+145)}{145^2-145}[/tex]
[tex]\displaystyle\longrightarrow\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{2\times145\times6}{145\times144}[/tex]
[tex]\displaystyle\longrightarrow\underline{\underline{\lim_{n\to\infty}\sum_{r=1}^n\alpha^r+\lim_{n\to\infty}\sum_{r=1}^n\beta^r=\dfrac{1}{12}}}[/tex]
Hence 1/12 is the answer.