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If a Cat has 4 kittens, and in 6 months those kittens have 4 kittens, and so on for 7 years, such that 14 batches of kittens are produced, from the one kitten. I produce the summation: $$\sum_{x=1}^{14} 4^x$$ Where added up looks like $4 + 16 + 64 + \cdots +4^{14}$.

Is there some kind of trick in maths for working this huge number out without a calculator?

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Look up en.wikipedia.org/wiki/Geometric_series. In your case, the sum is well approximated by its last member $4^{14} = 2^{28} \approx 200,000,000$ (using $2^{10} \approx 1000$). –  Yuval Filmus Feb 8 '11 at 1:53

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Let $\displaystyle S_x=\sum_{x=1}^{14}4^x=4+4^2+\ldots+4^{14}$ and then multiply both sides of $S_x$ by 4 to get $\displaystyle 4S_x=4^2+\ldots+4^{15}$. Now $4S_x-S_x=4^{15}-4$ so you can say that $\displaystyle S_x=\frac{4}{3}(4^{14}-1)$

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Comment, Cheers, well explained! –  DavidTheKiwiInSydney Mar 10 '11 at 4:57

Look up Geometric progression. The answer to your problem is $4 \times \frac{4^{14}-1}{3} = 357913940$

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