Question-29-Theres-2-same-ratio-sequences-a-n-b-n-with-a-ratio-of-non-zero-And-if-two-sums-of-each-sequences-n-1-a-n-n-1-b-n-are-convergent-and-two-equations-n-1-a-n-b

Question Number 213298 by MathematicalUser2357 last updated on 02/Nov/24

Q u e s t i o n 29 . T h e r e s 2 s a m e - r a t i o s e q u e n c e s {a_{n}}, {b_{n}} w i t h a r a t i o o f n o n - z e r o

A n d i f t w o s u m s o f e a c h s e q u e n c e s (\underset{n = 1}{\sum^{\infty}} a_{n}, \underset{n = 1}{\sum^{\infty}} b_{n}) a r e c o n v e r g e n t, a n d t w o e q u a t i o n s

\underset{n = 1}{\sum^{\infty}} a_{n} b_{n} = (\underset{n = 1}{\sum^{\infty}} a_{n}) \times (\underset{n = 1}{\sum^{\infty}} b_{n}) a n d 3 \times \underset{n = 1}{\sum^{\infty}} ∣ a_{2 n} ∣= 7 \times \underset{n = 1}{\sum^{\infty}} ∣ a_{3 n} ∣ a r e t r u e .

I f \underset{n = 1}{\sum^{\infty}} \frac{b_{2 n - 1} + b_{3 n + 1}}{b_{n}} = S, t h e n F i n d t h e v a l u e o f 120 S .

(k o r e a u n i v e r s i t y e x a m q u e s t i o n)

Answered by MrGaster last updated on 02/Nov/24

\underset{n = 1}{\sum^{\infty}} a_{n} = \frac{a_{1}}{1 - r_{a}}, \underset{n = 1}{\sum^{\infty}} b_{n} = \frac{b_{1}}{1 - r_{b}}

\underset{n = 1}{\sum^{\infty}} a_{n} b_{n} = \frac{a_{1} b_{1}}{1 - r_{a} r_{b}}

\frac{a_{1} b_{1}}{1 - r_{a} r_{b}} = \frac{a_{1}}{1 - r_{a}} \cdot \frac{b_{1}}{1 - r_{b}}

3 \underset{n = 1}{\sum^{\infty}} ∣ a_{2 n} ∣= 7 \underset{n = 1}{\sum^{\infty}} ∣ a_{3 n} ∣

3 \frac{∣ a_{1} ∣ r_{a}}{1 - r_{a}^{2}} = 7 \frac{∣ a_{1} ∣ r_{a}^{2}}{1 - r_{a}^{3}}

3 (1 - r_{a}^{3}) = 7 (1 - r_{a}^{2})

3 - 3 r_{a}^{3} - 7 r_{a}^{2} + 4 = 0

(r_{a} - 1) (3 r_{a}^{2} - 4 r_{a} - 4) = 0

r a = 1, r_{a} = 2, r_{a} = - \frac{2}{3}

r_{a} = 1 does not satisfy the condlition for

convergence . since ∣ r_{a} ∣< 1, have r_{a} = - \frac{2}{3} .

\underset{n = 1}{\sum^{\infty}} \frac{b_{2 n - 1} + b_{3 n + 1}}{b_{n}} = \underset{n = 1}{\sum^{\infty}} (\frac{b_{1} r_{b}^{2 n - 2} + b_{1} r_{b}^{3 n}}{b_{1} r_{b}^{n - 1}})

= \underset{n = 1}{\sum^{\infty}} (\frac{r_{b}^{2 n - 2} + b_{1} r_{b}^{3 n}}{r_{b}^{n - 1}})

= \underset{n = 1}{\sum^{\infty}} (r_{b}^{n - 1} + r_{b}^{2 n + 1})

= \underset{n = 1}{\sum^{\infty}} r_{b}^{n - 1} + \underset{n = 1}{\sum^{\infty}} r_{b}^{2 n + 1}

= \frac{1}{1 - r_{b}} + \frac{r_{b}^{3}}{(1 - r_{b}) (1 + r_{b})}

= \frac{1 + r_{b} + r_{b}^{3}}{1 - r_{b}^{2}}

= \frac{1 + r_{b} + r_{b}^{3}}{(1 - r_{b}) (1 + r_{b})}

= \frac{1 + r_{b} + r_{b}^{3}}{1 - r_{b}^{2}}

G iven that 120 S = 120 (\frac{1 + r_{b} + r_{b}^{3}}{1 - r_{b}^{2}}) . lf r_{b} = - \frac{2}{3}

120 S = 120 (\frac{1 - \frac{2}{3} + {(- \frac{2}{3})}^{3}}{1 - {(- \frac{2}{3})}^{2}})

= 120 (\frac{1 - \frac{2}{3} - \frac{8}{27}}{1 - \frac{4}{9}})

= 120 (\frac{27 - 18 - 8}{\frac{27}{\frac{5}{9}}})

= 120 (\frac{1}{\frac{27}{\frac{5}{8}}})

= 120 (\frac{1}{27} \cdot \frac{9}{5})

= 120 \cdot \frac{1}{15}

\begin{matrix} = 8 \end{matrix}

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