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Question Number 56411 by gunawan last updated on 16/Mar/19

For a sequence <a_n > ; a_1 =2, (a_(n+1) /a_n ) = (1/3),  then Σ_(r=1) ^(20)  a_r  is equal to

$$\mathrm{For}\:\mathrm{a}\:\mathrm{sequence}\:<{a}_{{n}} >\:;\:{a}_{\mathrm{1}} =\mathrm{2},\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}, \\ $$ $$\mathrm{then}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{20}} {\sum}}\:{a}_{{r}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19

(a_2 /a_1 )=(1/3)=(a_2 /2)→a_2 =(2/3)  (a_3 /a_2 )=(1/3)=(a_3 /(2/3))→a_3 =(2/3^2 )  (a_4 /a_3 )=(1/3)=(a_4 /(2/3^2 ))→a_4 =(2/3^3 )  S_n =a_1 +a_2 +a_3 +...+a_n   =(2/1)+(2/3)+(2/3^2 )+...+(2/3^(n−1) )  A=2   r=(1/3)    S_n =((2(1−(1/3^n )))/(1−(1/3)))→((2(1−(1/3^n )))/(2/3))→3(1−(1/3^n ))  so s_(20) =3(1−(1/3^(20) ))

$$\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{3}}=\frac{{a}_{\mathrm{2}} }{\mathrm{2}}\rightarrow{a}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$ $$\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}}=\frac{{a}_{\mathrm{3}} }{\frac{\mathrm{2}}{\mathrm{3}}}\rightarrow{a}_{\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} } \\ $$ $$\frac{{a}_{\mathrm{4}} }{{a}_{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}=\frac{{a}_{\mathrm{4}} }{\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }}\rightarrow{a}_{\mathrm{4}} =\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} } \\ $$ $${S}_{{n}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{{n}} \\ $$ $$=\frac{\mathrm{2}}{\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+...+\frac{\mathrm{2}}{\mathrm{3}^{{n}−\mathrm{1}} } \\ $$ $${A}=\mathrm{2}\:\:\:{r}=\frac{\mathrm{1}}{\mathrm{3}}\:\: \\ $$ $${S}_{{n}} =\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\rightarrow\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right)}{\frac{\mathrm{2}}{\mathrm{3}}}\rightarrow\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right) \\ $$ $${so}\:{s}_{\mathrm{20}} =\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{20}} }\right) \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

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