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макс228133744
@макс228133744
July 2022
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Найдите сумму первых: а) трех членов; б)шести членов геометрической прогрессии: 5; 5/6; ....
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AnonimusPro
Verified answer
A1=5; a2=5/6
q=a2/a1=)5/6)/5=1/6
a) S3=a1*(q^3-1)/q-1=a1*(q-1)(q^2+q+1)/q-1=a1*(q^2+q+1)=5*(1/36+1/6+1)=5*(1+6+36/36)=
5*43/36=215/36
S6=a1*(q^5-1)/q-1=(5*(1/6^5-1))/(-5/6)=-6(1/6^5-1)=6-(1/6^4)=6-1/1296=7775/1296
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Answers & Comments
Verified answer
A1=5; a2=5/6q=a2/a1=)5/6)/5=1/6
a) S3=a1*(q^3-1)/q-1=a1*(q-1)(q^2+q+1)/q-1=a1*(q^2+q+1)=5*(1/36+1/6+1)=5*(1+6+36/36)=
5*43/36=215/36
S6=a1*(q^5-1)/q-1=(5*(1/6^5-1))/(-5/6)=-6(1/6^5-1)=6-(1/6^4)=6-1/1296=7775/1296