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29 Dec 2019, 19:34
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B
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Difficulty:
95%
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Question Stats:
49% (02:34) correct
51%
(02:52)
wrong
based on 149
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Let \(a_n\) and \(b_n\) are the roots of the quadratic equation
\(nx^2 - x + n(n+1) = 0\), where n is positive integer.
Find the value of
\((\frac{1}{a_1}+\frac{1}{b_1})+(\frac{1}{a_2}+\frac{1}{b_2})+..........+(\frac{1}{a_{25}}+\frac{1}{b_{25}})\)
A. \(\frac{24}{25}\)
B. \(\frac{25}{26}\)
C. 1
D. \(\frac{25}{24}\)
E. \(\frac{26}{25}\)
\(nx^2 - x + n(n+1) = 0\), where n is positive integer.
Find the value of
\((\frac{1}{a_1}+\frac{1}{b_1})+(\frac{1}{a_2}+\frac{1}{b_2})+..........+(\frac{1}{a_{25}}+\frac{1}{b_{25}})\)
A. \(\frac{24}{25}\)
B. \(\frac{25}{26}\)
C. 1
D. \(\frac{25}{24}\)
E. \(\frac{26}{25}\)
29 Dec 2019, 20:57
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nick1816
In any quadratic equation the sum of roots is \(-\frac{b}{a}\)
Here in equation \(nx^2 - x + n(n+1) = 0\), b=-1 and a=n, so SUM= \(-\frac{b}{a}\)=\(-\frac{(-1)}{n}\)
Similarly Product of roots =\(\frac{c}{a}=\frac{n(n+1)}{n}\)
Here each pair \((\frac{1}{a_1}+\frac{1}{b_1})\) or \((\frac{1}{a_2}+\frac{1}{b_2})\) can be written as\( \frac{a_1+b_1}{a_1*b_1}\), that is sum of roots divided by product
So \(\frac{(\frac{1}{n})}{(\frac{n(n+1)}{n})}=\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{(n+1)}\), where \(1\leq n\leq 25\)
So the entire sequence becomes
\((\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3}).....(\frac{1}{24}-\frac{1}{25})+(\frac{1}{25}-\frac{1}{26})=1-\frac{1}{26}=\frac{25}{26}\)
B
General Discussion
16 Jul 2025, 20:40
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