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April 2021 1 6 Report

Найдите сумму первых шести членов геометрической прогрессии, если её десятый член равен 64, а знаменатель равен 1/2

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Радость25 $b_n=b_1*q^n^-^1 \\ \\ b_1= \frac{b_n}{q^n^-^1}= \frac{64}{ (\frac{1}{2})^9 } =64*512=32768 \\ \\ S_6= \frac{b_1*(q^6-1)}{q-1} =\frac{32768*((1/2)^6-1)}{1/2-1} =\frac{32768*(1/64-1)}{-1/2}=\frac{32768*(-63/64)}{-1/2}= \\ =64512$

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egor gotovilsya k ekzamenu po geometrii za 5 dnej emu nado bylo vyuchit 30 bilet

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