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Question Number 78460 by Tony Lin last updated on 17/Jan/20
Σ_(n=1) ^∞ (n^3 /3^n )=?
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }=? \\ $$
Commented by mathmax by abdo last updated on 18/Jan/20
let s =Σ_(n=1) ^∞ (n^3 /3^n ) and f(x)=Σ_(n=0) ^∞ x^n with ∣x∣<1 ⇒f(x)=(1/(1−x)) ⇒f^((1)) (x)=(1/((1−x)^2 )) =Σ_(n=1) ^∞ nx^(n−1) ⇒(x/((x−1)^2 )) =Σ_(n=1) ^∞ nx^n ⇒ Σ_(n=1) ^∞ n^2 x^(n−1) =((x/((x−1)^2 )))^((1)) =(((x−1)^2 −2(x−1)x)/((x−1)^4 )) =((x−1−2x)/((x−1)^3 )) =((−x−1)/((x−1)^3 )) =((x+1)/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ n^2 x^n =((x^2 +x)/((1−x)^3 )) ⇒Σ_(n=1) ^∞ n^3 x^(n−1) =(((x^2 +x)/((1−x)^3 )))^((1)) =(((2x+1)(1−x)^3 +3(1−x)^2 (x^2 +x))/((1−x)^6 )) =(((2x+1)(1−x)+3(x^2 +x))/((1−x)^4 )) =((2x−2x^2 +1−x +3x^2 +3x)/((1−x)^4 )) =((x^2 +4x+1)/((1−x)^4 )) ⇒Σ_(n=1) ^∞ n^3 x^n =((x^3 +4x^2 +x)/((1−x)^4 )) ⇒s =f((1/3)) =((((1/3))^3 +4((1/3))^2 +(1/3))/((1−(1/3))^4 )) =(((1/3^3 )+(4/3^2 )+(1/3))/(2^4 /3^4 )) =((3^4 {(1/3^3 )+(4/3^2 )+(1/3)})/(16)) =((3+36 +27)/(16)) =((39+27)/(16)) =((66)/(16)) =((33)/8) ★ s=((33)/8)★
$${let}\:{s}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }\:\:{and}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} \:\:{with}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{f}^{\left(\mathrm{1}\right)} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}−\mathrm{1}} \:\Rightarrow\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} =\left(\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\right)^{\left(\mathrm{1}\right)} =\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}−\mathrm{1}\right){x}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}−\mathrm{1}−\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{−{x}−\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}} \:=\frac{{x}^{\mathrm{2}} +{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{\mathrm{3}} {x}^{{n}−\mathrm{1}} =\left(\frac{{x}^{\mathrm{2}} +{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\right)^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{6}} }\:=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)+\mathrm{3}\left({x}^{\mathrm{2}} +{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}−{x}\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{3}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{\mathrm{3}} {x}^{{n}} =\frac{{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow{s}\:={f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} \:+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{4}} }\:=\frac{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{4}} }} \\ $$$$=\frac{\mathrm{3}^{\mathrm{4}} \left\{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}\right\}}{\mathrm{16}}\:=\frac{\mathrm{3}+\mathrm{36}\:+\mathrm{27}}{\mathrm{16}}\:=\frac{\mathrm{39}+\mathrm{27}}{\mathrm{16}}\:=\frac{\mathrm{66}}{\mathrm{16}}\:=\frac{\mathrm{33}}{\mathrm{8}} \\ $$$$\bigstar\:{s}=\frac{\mathrm{33}}{\mathrm{8}}\bigstar \\ $$
Commented by Tony Lin last updated on 18/Jan/20
thanks sir
$${thanks}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 18/Jan/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$
Answered by mr W last updated on 17/Jan/20
1+x+x^2 +x^3 +...=(1/(1−x)) with ∣x∣<1 1+2x+3x^2 +4x^3 +...=(1/((1−x)^2 )) x+2x^2 +3x^3 +4x^4 +...=(x/((1−x)^2 )) 1+2^2 x+3^2 x^2 +4^2 x^3 +...=(1/((1−x)^2 ))+((2x)/((1−x)^3 )) x+2^2 x^2 +3^2 x^3 +4^2 x^4 +...=(x/((1−x)^2 ))+((2x^2 )/((1−x)^3 )) 1+2^3 x+3^3 x^2 +4^3 x^3 +...=(1/((1−x)^2 ))+((6x)/((1−x)^3 ))+((6x^2 )/((1−x)^4 )) x+2^3 x^2 +3^3 x^3 +4^3 x^4 +...=(x/((1−x)^2 ))+((6x^2 )/((1−x)^3 ))+((6x^3 )/((1−x)^4 )) let x=(1/3): (1^3 /3)+(2^3 /3^2 )+(3^3 /3^3 )+(4^3 /3^4 )+...=Σ_(n=1) ^∞ (n^3 /3^n )=(1/3)×(3^2 /2^2 )+(6/3^2 )×(3^3 /2^3 )+(6/3^3 )×(3^4 /2^4 ) ⇒Σ_(n=1) ^∞ (n^3 /3^n )=((33)/8)
$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +…=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +…=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{1}+\mathrm{2}^{\mathrm{2}} {x}+\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} {x}^{\mathrm{3}} +…=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$${x}+\mathrm{2}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{4}^{\mathrm{2}} {x}^{\mathrm{4}} +…=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$\mathrm{1}+\mathrm{2}^{\mathrm{3}} {x}+\mathrm{3}^{\mathrm{3}} {x}^{\mathrm{2}} +\mathrm{4}^{\mathrm{3}} {x}^{\mathrm{3}} +…=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{6}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }+\frac{\mathrm{6}{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$${x}+\mathrm{2}^{\mathrm{3}} {x}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} {x}^{\mathrm{4}} +…=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{6}{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }+\frac{\mathrm{6}{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$${let}\:{x}=\frac{\mathrm{1}}{\mathrm{3}}: \\ $$$$\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }+…=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{3}} }×\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} } \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }=\frac{\mathrm{33}}{\mathrm{8}} \\ $$
Commented by Tony Lin last updated on 18/Jan/20
thanks sir
$${thanks}\:{sir} \\ $$
Answered by mind is power last updated on 17/Jan/20
x^n .n^3 n(n−1)(n−2)=n^3 −3n^2 +2n n^3 =n(n−1)(n−2)+n(3n−2)=n(n−1)(n−2)+3n(n−1)+n ⇒Σ_(n≥1) n(n−1)(n−2)x^n +3Σ_(n≥1) n(n−1)x^n +Σ_(n≥1) nx^n =x^3 Σ_(n≥2) n(n−1)(n−2)x^(n−3) +3x^2 Σ_(n≥2) n(n−1)x^(n−2) +xΣ_(n≥1) nx^(n−1) Σ_(n≥0) x^n =(1/(1−x)) ⇒Σ_(n≥1) nx^(n−1) =(1/((1−x)^2 ))⇒Σn(n−1)(n−2)x^(n−3) =(6/((1−x)^4 )) Σn(n−1)x^(n−2) =(2/((1−x)^3 )) =((6x^3 )/((x−1)^4 ))+((6x^2 )/((1−x)^3 ))+(x/((1−x)^2 )) x=3⇒((6.27)/(16))+((6.9)/(−8))+(3/4)=((162−108+12)/(16))=((66)/(16))=((33)/8)
$$\mathrm{x}^{\mathrm{n}} .\mathrm{n}^{\mathrm{3}} \\ $$$$\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)=\mathrm{n}^{\mathrm{3}} −\mathrm{3n}^{\mathrm{2}} +\mathrm{2n} \\ $$$$\mathrm{n}^{\mathrm{3}} =\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)+\mathrm{n}\left(\mathrm{3n}−\mathrm{2}\right)=\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)+\mathrm{3n}\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{n} \\ $$$$\Rightarrow\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}} +\mathrm{3}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}} +\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{nx}^{\mathrm{n}} \\ $$$$=\mathrm{x}^{\mathrm{3}} \underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}−\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} \underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{x}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{nx}^{\mathrm{n}−\mathrm{1}} \\ $$$$\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\mathrm{x}^{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}} \\ $$$$\Rightarrow\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{nx}^{\mathrm{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\Rightarrow\Sigma\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)\mathrm{x}^{\mathrm{n}−\mathrm{3}} =\frac{\mathrm{6}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} } \\ $$$$\Sigma\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{2}} =\frac{\mathrm{2}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{6x}^{\mathrm{3}} }{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{6x}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{3}} }+\frac{\mathrm{x}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{x}=\mathrm{3}\Rightarrow\frac{\mathrm{6}.\mathrm{27}}{\mathrm{16}}+\frac{\mathrm{6}.\mathrm{9}}{−\mathrm{8}}+\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{162}−\mathrm{108}+\mathrm{12}}{\mathrm{16}}=\frac{\mathrm{66}}{\mathrm{16}}=\frac{\mathrm{33}}{\mathrm{8}} \\ $$
Commented by Tony Lin last updated on 18/Jan/20
thanks sir
$${thanks}\:{sir} \\ $$
Commented by mind is power last updated on 18/Jan/20
y′re welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$
Answered by john santu last updated on 18/Jan/20
let S= (1/3)+(8/9)+((27)/(27))+((64)/(81))+((125)/(243))+... (1) by multiply in both side with (1/3) (1/3)S= (1/9)+(8/(27))+((27)/(64))+((64)/(243))+... (2) substract equations 1 and 2 (2/3)S=(1/3)+(7/9)+((19)/(27))+((37)/(81))+((61)/(243))+...(3) (2/9)S= (1/9)+(7/(27))+((19)/(81))+((37)/(243))+... (4/9)S=(1/3)+(6/9)+((12)/(27))+((18)/(81))+((24)/(243))+... (4/9)S= (1/3)+6((1/9)+(2/(27))+(3/(81))+(4/(243))+...) let P= (1/9)+(2/(27))+(3/(81))+(4/(243))+... (1/3)P= (1/(27))+(2/(81))+(3/(243))+... (2/3)P= (1/9)+(1/(27))+(1/(81))+(1/(243))+...=(1/6) ⇒P= (1/4) . now we get (4/9)S= (1/3)+6×(1/4) ⇒S= ((33)/(8 )) .★
$${let}\:{S}=\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{9}}+\frac{\mathrm{27}}{\mathrm{27}}+\frac{\mathrm{64}}{\mathrm{81}}+\frac{\mathrm{125}}{\mathrm{243}}+…\:\left(\mathrm{1}\right) \\ $$$${by}\:{multiply}\:{in}\:{both}\:{side}\:{with}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}{S}=\:\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{8}}{\mathrm{27}}+\frac{\mathrm{27}}{\mathrm{64}}+\frac{\mathrm{64}}{\mathrm{243}}+…\:\left(\mathrm{2}\right) \\ $$$${substract}\:{equations}\:\mathrm{1}\:{and}\:\mathrm{2} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{S}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{7}}{\mathrm{9}}+\frac{\mathrm{19}}{\mathrm{27}}+\frac{\mathrm{37}}{\mathrm{81}}+\frac{\mathrm{61}}{\mathrm{243}}+…\left(\mathrm{3}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}{S}=\:\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{7}}{\mathrm{27}}+\frac{\mathrm{19}}{\mathrm{81}}+\frac{\mathrm{37}}{\mathrm{243}}+… \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}{S}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{6}}{\mathrm{9}}+\frac{\mathrm{12}}{\mathrm{27}}+\frac{\mathrm{18}}{\mathrm{81}}+\frac{\mathrm{24}}{\mathrm{243}}+… \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}{S}=\:\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{27}}+\frac{\mathrm{3}}{\mathrm{81}}+\frac{\mathrm{4}}{\mathrm{243}}+…\right) \\ $$$${let}\:{P}=\:\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{27}}+\frac{\mathrm{3}}{\mathrm{81}}+\frac{\mathrm{4}}{\mathrm{243}}+… \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}{P}=\:\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{81}}+\frac{\mathrm{3}}{\mathrm{243}}+… \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{P}=\:\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{1}}{\mathrm{81}}+\frac{\mathrm{1}}{\mathrm{243}}+…=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{P}=\:\frac{\mathrm{1}}{\mathrm{4}}\:.\:{now}\:{we}\:{get}\: \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}{S}=\:\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{6}×\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{S}=\:\frac{\mathrm{33}}{\mathrm{8}\:}\:.\bigstar \\ $$
Commented by Tony Lin last updated on 18/Jan/20
thanks sir
$${thanks}\:{sir} \\ $$

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