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10-25-28-125-82-625-




Question Number 143184 by bramlexs22 last updated on 11/Jun/21
    ((10)/(25))+((28)/(125))+((82)/(625))+... = ?
$$\:\:\:\:\frac{\mathrm{10}}{\mathrm{25}}+\frac{\mathrm{28}}{\mathrm{125}}+\frac{\mathrm{82}}{\mathrm{625}}+…\:=\:? \\ $$
Answered by Canebulok last updated on 11/Jun/21
Solution:  In terms of summation,  ⇒ Σ_(n=1) ^∞  (((3^(n+1) ) + 1)/5^(n+1) ) =  Σ_(n=1) ^∞  ((3/5))^(n+1) +  Σ_(n=1) ^∞   ((1/5))^(n+1)       By calculus method,  ⇒ Σ_(n=0) ^∞   x^n  = (1/((1−x)))     By shifting the index of summation,  → Let n = k−1  ⇒ Σ_(k=1) ^∞  x^((k−1))  =  (1/((1−x)))     By multiplying both sides by “ x^2  ”,  ⇒ Σ_(k=1) ^∞  x^((k+1))  =  (x^2 /((1−x)))     Let x = (1/5)     ⇒ Σ_(n=1) ^∞   ((1/5))^(n+1)  = ((((1/5))^2 )/((1−(1/5))))  =  (1/(20))     Let x = (3/5)     ⇒ Σ_(n=1) ^∞   ((3/5))^(n+1)  =  ((((3/5))^2 )/((1−(3/5))))  =  (9/(10))     Thus;  ⇒ Σ_(n=1) ^∞   (((3^(n+1) ) + 1)/5^(n+1) ) =  (1/(20)) + (9/(10))  =  ((19)/(20))     ∼ Kevin
$$\boldsymbol{{Solution}}: \\ $$$${In}\:{terms}\:{of}\:{summation}, \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{3}^{{n}+\mathrm{1}} \right)\:+\:\mathrm{1}}{\mathrm{5}^{{n}+\mathrm{1}} }\:=\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}+\mathrm{1}} +\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}+\mathrm{1}} \: \\ $$$$\: \\ $$$${By}\:{calculus}\:{method}, \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:{x}^{{n}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)} \\ $$$$\: \\ $$$${By}\:{shifting}\:{the}\:{index}\:{of}\:{summation}, \\ $$$$\rightarrow\:{Let}\:{n}\:=\:{k}−\mathrm{1} \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:{x}^{\left({k}−\mathrm{1}\right)} \:=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)} \\ $$$$\: \\ $$$${By}\:{multiplying}\:{both}\:{sides}\:{by}\:“\:{x}^{\mathrm{2}} \:'', \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:{x}^{\left({k}+\mathrm{1}\right)} \:=\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)} \\ $$$$\: \\ $$$${Let}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\: \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}+\mathrm{1}} \:=\:\frac{\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{20}} \\ $$$$\: \\ $$$${Let}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\: \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}+\mathrm{1}} \:=\:\:\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}\right)}\:\:=\:\:\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$\: \\ $$$${Thus}; \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\left(\mathrm{3}^{{n}+\mathrm{1}} \right)\:+\:\mathrm{1}}{\mathrm{5}^{{n}+\mathrm{1}} }\:=\:\:\frac{\mathrm{1}}{\mathrm{20}}\:+\:\frac{\mathrm{9}}{\mathrm{10}}\:\:=\:\:\frac{\mathrm{19}}{\mathrm{20}} \\ $$$$\: \\ $$$$\sim\:{Kevin} \\ $$

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