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Valentin [98]
3 years ago
10

Consider the infinite geometric series (∞/Σ/n=1) * -4(1/3)^{n-1}.

Mathematics
1 answer:
raketka [301]3 years ago
5 0
The first four terms of the series probably refer to the first four partial sums.

\displaystyle\sum_{n=1}^1-4\left(\frac13\right)^{n-1}=-4\left(\frac13\right)^{1-1}=-4
\displaystyle\sum_{n=1}^2-4\left(\frac13\right)^{n-1}=-\frac{16}3
\displaystyle\sum_{n=1}^3-4\left(\frac13\right)^{n-1}=-\frac{52}9
\displaystyle\sum_{n=1}^4-4\left(\frac13\right)^{n-1}=-\frac{160}{27}

which you can compute either by adding one term at a time, or using the well-known formula,

\displaystyle\sum_{n=1}^kar^{n-1}=a\frac{1-r^{k+1}}{1-r}

(I can provide a link to a derivation I gave in a nearly identical question in the comments)

The series converges as k\to\infty if and only if |r|, which is certainly the case here, since -1.

Extrapolating from the formula above, the sum of the convergent series is

\displaystyle\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}

so the sum for this series is \dfrac{-4}{1-\frac13}=-6.
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