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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

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Question Number 213298 by MathematicalUser2357 last updated on 02/Nov/24
Question 29. Theres 2 same-ratio sequences {a_n },{b_n } with a ratio of non-zero And if two sums of each sequences (Σ_(n=1) ^∞ a_n ,Σ_(n=1) ^∞ b_n ) are convergent, and two equations Σ_(n=1) ^∞ a_n b_n =(Σ_(n=1) ^∞ a_n )×(Σ_(n=1) ^∞ b_n ) and 3×Σ_(n=1) ^∞ ∣a_(2n) ∣=7×Σ_(n=1) ^∞ ∣a_(3n) ∣ are true. If Σ_(n=1) ^∞ ((b_(2n−1) +b_(3n+1) )/b_n )=S, then Find the value of 120S. (korea university exam question)
$$\boldsymbol{{Question}}\:\mathrm{29}.\:{Theres}\:\mathrm{2}\:{same}-{ratio}\:{sequences}\:\left\{{a}_{{n}} \right\},\left\{{b}_{{n}} \right\}\:{with}\:{a}\:{ratio}\:{of}\:{non}-{zero} \\ $$$${And}\:{if}\:{two}\:{sums}\:{of}\:{each}\:{sequences}\:\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ,\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} \right)\:{are}\:{convergent},\:{and}\:{two}\:{equations} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} {b}_{{n}} =\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \right)×\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} \right)\:{and}\:\mathrm{3}×\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{a}_{\mathrm{2}{n}} \mid=\mathrm{7}×\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{a}_{\mathrm{3}{n}} \mid\:{are}\:{true}. \\ $$$${If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{b}_{\mathrm{2}{n}−\mathrm{1}} +{b}_{\mathrm{3}{n}+\mathrm{1}} }{{b}_{{n}} }={S},\:{then}\:\boldsymbol{{Find}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\mathrm{120}\boldsymbol{{S}}. \\ $$$$\left({korea}\:{university}\:{exam}\:{question}\right) \\ $$
Answered by MrGaster last updated on 02/Nov/24
Σ_(n=1) ^∞ a_n =(a_1 /(1−r_a )),Σ_(n=1) ^∞ b_n =(b_1 /(1−r_b )) Σ_(n=1) ^∞ a_n b_n =((a_1 b_1 )/(1−r_a r_b )) ((a_1 b_1 )/(1−r_a r_b ))=(a_1 /(1−r_a ))∙(b_1 /(1−r_b )) 3Σ_(n=1) ^∞ ∣a_(2n) ∣=7Σ_(n=1) ^∞ ∣a_(3n) ∣ 3((∣a_1 ∣r_a )/(1−r_a ^2 ))=7((∣a_1 ∣r_a ^2 )/(1−r_a ^3 )) 3(1−r_a ^3 )=7(1−r_a ^2 ) 3−3r_a ^3 −7r_a ^2 +4=0 (r_a −1)(3r_a ^2 −4r_a −4)=0 ra=1,r_a =2,r_a =−(2/3) r_a =1 does not satisfy the condlition for convergence.since∣r_a ∣<1, have r_a =−(2/3). Σ_(n=1) ^∞ ((b_(2n−1) +b_(3n+1) )/b_n )=Σ_(n=1) ^∞ (((b_1 r_b ^(2n−2) +b_1 r_b ^(3n) )/(b_1 r_b ^(n−1) ))) =Σ_(n=1) ^∞ (((r_b ^(2n−2) +b_1 r_b ^(3n) )/r_b ^(n−1) )) =Σ_(n=1) ^∞ (r_b ^(n−1) +r_b ^(2n+1) ) =Σ_(n=1) ^∞ r_b ^(n−1) +Σ_(n=1) ^∞ r_b ^(2n+1) =(1/(1−r_b ))+(r_b ^3 /((1−r_b )(1+r_b ))) =((1+r_b +r_b ^3 )/(1−r_b ^2 )) =((1+r_b +r_b ^3 )/((1−r_b )(1+r_b ))) =((1+r_b +r_b ^3 )/(1−r_b ^2 )) Given that 120S=120(((1+r_b +r_b ^3 )/(1−r_b ^2 ))).lf r_b =−(2/3) 120S=120(((1−(2/3)+(−(2/3))^3 )/(1−(−(2/3))^2 ))) =120(((1−(2/3)−(8/(27)))/(1−(4/9)))) =120(((27−18−8)/((27)/(5/9)))) =120((1/((27)/(5/8)))) =120((1/(27))∙(9/5)) =120∙(1/(15)) determinant (((=8)))
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} =\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{r}_{{a}} },\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} =\frac{{b}_{\mathrm{1}} }{\mathrm{1}−{r}_{{b}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} {b}_{{n}} =\frac{{a}_{\mathrm{1}} {b}_{\mathrm{1}} }{\mathrm{1}−{r}_{{a}} {r}_{{b}} } \\ $$$$\frac{{a}_{\mathrm{1}} {b}_{\mathrm{1}} }{\mathrm{1}−{r}_{{a}} {r}_{{b}} }=\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{r}_{{a}} }\centerdot\frac{{b}_{\mathrm{1}} }{\mathrm{1}−{r}_{{b}} } \\ $$$$\mathrm{3}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{a}_{\mathrm{2}{n}} \mid=\mathrm{7}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{a}_{\mathrm{3}{n}} \mid \\ $$$$\mathrm{3}\frac{\mid{a}_{\mathrm{1}} \mid{r}_{{a}} }{\mathrm{1}−{r}_{{a}} ^{\mathrm{2}} }=\mathrm{7}\frac{\mid{a}_{\mathrm{1}} \mid{r}_{{a}} ^{\mathrm{2}} }{\mathrm{1}−{r}_{{a}} ^{\mathrm{3}} } \\ $$$$\mathrm{3}\left(\mathrm{1}−{r}_{{a}} ^{\mathrm{3}} \right)=\mathrm{7}\left(\mathrm{1}−{r}_{{a}} ^{\mathrm{2}} \right) \\ $$$$\mathrm{3}−\mathrm{3}{r}_{{a}} ^{\mathrm{3}} −\mathrm{7}{r}_{{a}} ^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\left({r}_{{a}} −\mathrm{1}\right)\left(\mathrm{3}{r}_{{a}} ^{\mathrm{2}} −\mathrm{4}{r}_{{a}} −\mathrm{4}\right)=\mathrm{0} \\ $$$${ra}=\mathrm{1},{r}_{{a}} =\mathrm{2},{r}_{{a}} =−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${r}_{{a}} =\mathrm{1}\:\mathrm{does}\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{condlition}\:\mathrm{for} \\ $$$$\mathrm{convergence}.\mathrm{since}\mid{r}_{{a}} \mid<\mathrm{1},\:\mathrm{have}\:\:{r}_{{a}} =−\frac{\mathrm{2}}{\mathrm{3}}. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{b}_{\mathrm{2}{n}−\mathrm{1}} +{b}_{\mathrm{3}{n}+\mathrm{1}} }{{b}_{{n}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{b}_{\mathrm{1}} {r}_{{b}} ^{\mathrm{2}{n}−\mathrm{2}} +{b}_{\mathrm{1}} {r}_{{b}} ^{\mathrm{3}{n}} }{{b}_{\mathrm{1}} {r}_{{b}} ^{{n}−\mathrm{1}} }\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{r}_{{b}} ^{\mathrm{2}{n}−\mathrm{2}} +{b}_{\mathrm{1}} {r}_{{b}} ^{\mathrm{3}{n}} }{{r}_{{b}} ^{{n}−\mathrm{1}} }\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({r}_{{b}} ^{{n}−\mathrm{1}} +{r}_{{b}} ^{\mathrm{2}{n}+\mathrm{1}} \right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{r}_{{b}} ^{{n}−\mathrm{1}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{r}_{{b}} ^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{r}_{{b}} }+\frac{{r}_{{b}} ^{\mathrm{3}} }{\left(\mathrm{1}−{r}_{{b}} \right)\left(\mathrm{1}+{r}_{{b}} \right)} \\ $$$$=\frac{\mathrm{1}+{r}_{{b}} +{r}_{{b}} ^{\mathrm{3}} }{\mathrm{1}−{r}_{{b}} ^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}+{r}_{{b}} +{r}_{{b}} ^{\mathrm{3}} }{\left(\mathrm{1}−{r}_{{b}} \right)\left(\mathrm{1}+{r}_{{b}} \right)} \\ $$$$=\frac{\mathrm{1}+{r}_{{b}} +{r}_{{b}} ^{\mathrm{3}} }{\mathrm{1}−{r}_{{b}} ^{\mathrm{2}} } \\ $$$${G}\mathrm{iven}\:\mathrm{that}\:\mathrm{120}{S}=\mathrm{120}\left(\frac{\mathrm{1}+{r}_{{b}} +{r}_{{b}} ^{\mathrm{3}} }{\mathrm{1}−{r}_{{b}} ^{\mathrm{2}} }\right).\mathrm{lf}\:{r}_{{b}} =−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{120}{S}=\mathrm{120}\left(\frac{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}+\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }{\mathrm{1}−\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }\right) \\ $$$$=\mathrm{120}\left(\frac{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{27}}}{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}}\right) \\ $$$$=\mathrm{120}\left(\frac{\mathrm{27}−\mathrm{18}−\mathrm{8}}{\frac{\mathrm{27}}{\frac{\mathrm{5}}{\mathrm{9}}}}\right) \\ $$$$=\mathrm{120}\left(\frac{\mathrm{1}}{\frac{\mathrm{27}}{\frac{\mathrm{5}}{\mathrm{8}}}}\right) \\ $$$$=\mathrm{120}\left(\frac{\mathrm{1}}{\mathrm{27}}\centerdot\frac{\mathrm{9}}{\mathrm{5}}\right) \\ $$$$=\mathrm{120}\centerdot\frac{\mathrm{1}}{\mathrm{15}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{=\mathrm{8}}\\\hline\end{array} \\ $$

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