3 years ago

Evaluate.

Mathematics

1 answer:

marin [14]3 years ago

7 0

3-7*5+2*4
3-35+8
-32+8
-24
B

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We can employ a simple repeated decimal trick:

$x=1.111\ldots$

$0.1x=0.111\ldots$

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###

Alternatively, we can compute the partial sum of the series.

$\displaystyle S_n=\sum_{k=0}^n\dfrac1{10^k}$

$S_n=1+0.1+0.01+\cdots+\dfrac1{10^n}$

$0.1S_n=0.1+0.01+0.001+\cdots+\dfrac1{10^{n+1}}$

$\implies S_n-0.1S_n=0.9S_n=1-\dfrac1{10^{n+1}}$

$\implies S_n=\dfrac{10}9-\dfrac9{10^n}$

As $n\to\infty$ , the second term vanishes and we're left with $\dfrac{10}9$ . Notice that this is really just a more formal version of the earlier trick.

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