How do you find the number of terms given $s_{n} = 3829$ $s_n=3829$ and $7+(-21)+63+(-189)+...$ ?

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Answer 1 · 2016-08-21T21:57:28

$n = 6$

$7+(-21)+63+(-189)+cdots = 7sum_{k=0}^{k=n}(-3)^k = S_n$

We know that

$sum_{k=0}^{k=n}z^k = (z^{n+1}-1)/(z-1)$

so

$S_n = 7((-3)^{n+1}-1)/(-3-1) = 3829$

then

$(-3)^{n+1}=(-4)3829/7+1$

and

$(-3)^n = (4 cdot 3829/7-1)/3 = 729 = 3^6$

so

$n = 6$

How do you find the number of terms given s_n=3829sn=3829s_n=3829 and 7+(-21)+63+(-189)+...7+(-21)+63+(-189)+...?