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Prove that:$${3}^{{\frac{{1}}{{2}}}}\times{3}^{{\frac{{1}}{{4}}}}\times{3}^{{\frac{{1}}{{8}}}}\times\ldots={3}$$
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## QuestionMathsClass 11

Prove that:$${3}^{{\frac{{1}}{{2}}}}\times{3}^{{\frac{{1}}{{4}}}}\times{3}^{{\frac{{1}}{{8}}}}\times\ldots={3}$$

$${L}.{H}.{S}.={3}^{{\frac{{1}}{{2}}}}\times{3}^{{\frac{{1}}{{4}}}}\times{3}^{{\frac{{1}}{{8}}}}+\ldots$$

$$={3}^{{\frac{{1}}{{2}}+\frac{{1}}{{4}}+\frac{{1}}{{8}}+\frac{{1}}{{16}}\ldots}}$$

Now, we will solve, $${\left(\frac{{1}}{{2}}+\frac{{1}}{{4}}+\frac{{1}}{{8}}+\frac{{1}}{{16}}\ldots\right)}$$

It forms a GP with first term, $${a}=\frac{{1}}{{2}}$$ and common ratio, $${r}=\frac{{1}}{{2}}$$

So, Sum of the given series will be, $${S}=\frac{{a}}{{{1}-{r}}}=\frac{{\frac{{1}}{{2}}}}{{{1}-\frac{{1}}{{2}}}}={1}$$

So, our expression becomes,

$${3}^{{\frac{{1}}{{2}}+\frac{{1}}{{4}}+\frac{{1}}{{8}}\ldots}}={3}^{{1}}={3}={R}.{H}.{S}.$$