# Further simplify this $\sum \limits_{i=1}^{k}{{k \choose i} \cdot 12^i \cdot 2^i}$

Could we further simplify this:

$$\sum_{i=1}^{k}{{k \choose i} \cdot 12^i \cdot 2^i}$$

or, at least, find a close upper bound?

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Firstly, note that $12^i * 2^i$ is nothing more than $24^i$.

So consider $1 + \displaystyle\sum_{i=1}^{k}{{k \choose i} \times 12^i \times 2^i} = \displaystyle\sum_{i=0}^{k}{{k \choose i} \times 24^i} = (1 + 24)^k$. (by the binomial theorem).

So your sum is nothing more than $25^k - 1$.

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Great, thanks also for the explanation! –  Asterios Jun 6 '11 at 13:35
keeping in mind that $12^i\times 2^i$ is just $(12\times 2)^i$ ;) –  A. De Luca Jun 6 '11 at 13:22
$(1+24)^k=\sum\limits_{i = 0}^k {{k \choose i}24^i }$.