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Question-125641




Question Number 125641 by aurpeyz last updated on 12/Dec/20
Answered by Dwaipayan Shikari last updated on 12/Dec/20
S=2+(5/2)+(8/2^2 )+((11)/2^3 )+...  (S/2)=      (2/2)+(5/2^2 )+(8/2^3 )+..  S−(S/2)=2+(3/2)+(3/2^2 )+...⇒S=4+6(((1/2)/(1−(1/2))))=10
S=2+52+822+1123+S2=22+522+823+..SS2=2+32+322+S=4+6(12112)=10
Commented by aurpeyz last updated on 12/Dec/20
thanks
thanks
Answered by mathmax by abdo last updated on 12/Dec/20
S=(2/2^0 )+(5/2^1 )+(8/2^2 )+...=Σ_(n=0) ^∞  (a_n /2^n )  let determine a_n   a_n =αn +β?    we have a_1 =5=α+β  a_2 =8 =2α +β ⇒ { ((α+β=5)),((2α+β=8 ⇒α=3  and β=2 ⇒a_n =3n+2)) :}  a_0 =2 ⇒ S =Σ_(n=0) ^∞  ((3n+2)/2^n ) =3 Σ_(n=0) ^∞  (n/2^n ) +2 Σ_(n=0) ^∞  (1/2^n ) we have  Σ_(n=0) ^∞  (1/2^n )=(1/(1−(1/2)))=2  and Σ_(n=0) ^∞ (n/2^n )=s((1/2))with s(x)=Σ_(n=1) ^∞  nx^n   Σ_(n=0) ^∞  x^n  =(1/(1−x)) ⇒Σ_(n=1) ^∞  nx^(n−1)  =(1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^n  =(x/((1−x)^2 ))  =s(x) ⇒s((1/2))=(1/(2(1−(1/2))^2 ))=(1/(2×(1/4)))=2 ⇒  S=3.2 +4 =6+4 =10
S=220+521+822+=n=0an2nletdetermineanan=αn+β?wehavea1=5=α+βa2=8=2α+β{α+β=52α+β=8α=3andβ=2an=3n+2a0=2S=n=03n+22n=3n=0n2n+2n=012nwehaven=012n=1112=2andn=0n2n=s(12)withs(x)=n=1nxnn=0xn=11xn=1nxn1=1(1x)2n=1nxn=x(1x)2=s(x)s(12)=12(112)2=12×14=2S=3.2+4=6+4=10
Commented by aurpeyz last updated on 13/Dec/20
those equations in line 6 are laws?
thoseequationsinline6arelaws?

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