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Find the sum of $\binom{100}1 + 2\binom{100}2 + 4\binom{100}3 +8\binom{100}4+\dots+2^{99}\binom{100}{100}$

How you guys work on with this question? With the geometric progression? Combination? Or anyother way to calculate?

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  • $\begingroup$ Have you tried expanding $(1 + 2)^{100}$ using the binomial theorem? $\endgroup$ – Arthur Mar 3 '14 at 16:04
  • $\begingroup$ There was a misplaced parenthesis in the original question. Please check that the latexification was done right. $\endgroup$ – Daniel R Mar 3 '14 at 16:13
  • $\begingroup$ Thx for help in edit my question.... $\endgroup$ – user132564 Mar 3 '14 at 16:25
  • $\begingroup$ It looks like you have some extra parentheses. $\endgroup$ – Ross Millikan Mar 3 '14 at 16:33
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$$\sum_{r=1}^{100}2^{r-1}\binom{100}r=\frac12\sum_{r=1}^{100}2^r\binom{100}r=\frac12\left[(1+2)^{100}-1\right]$$

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