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Prove for a medal.
The sum of the first \(n\) terms of the geometric sequence \(S_n\).
\(b_1\) is the first term, \(q\) is the ratio.
 one year ago
 one year ago
Prove for a medal. The sum of the first \(n\) terms of the geometric sequence \(S_n\). \(b_1\) is the first term, \(q\) is the ratio.
 one year ago
 one year ago

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klimenkovBest ResponseYou've already chosen the best response.0
Please, write a formula for \(S_n\) and it's proof.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
It is better, you don't use any sourses to help yourself.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
if \(b_n=b_1q^{n1}\). Let \[ \large S_n=b_1+b_2++b_3\dots+b_n=b_1+b_1q+b_1q^2+\dots+b_1q^{n1} \] then \[ \large qS_n=b_1q+b_1q^2+\dots+b_1q^n \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1349108015779:dw
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
from these \[ \large qS_nS_n=b_1q^nb_1 \] and if \(q\neq1\) \[ \large S_n=b_1\cdot\frac{q^n1}{q1} \]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
So, who owns a medal? Who was really the first? I think @helder_edwin.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
i don't think that matters.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.2
i think the point is that if u found the posts useful.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Yes, you are right. Hope @experimentX won't be sad about this.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
doesn't matter much to me
 one year ago
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