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I have some problem to prove this statement by the Principle of mathematical Induction.

$$\sum_{i=0}^{n} \binom{n}{i} = 2^n.$$ So I begin with the base step. For $n=0$, $$\sum_{i=0}^{0} \binom{0}{i} = 2^0 =1.$$

Now could you help me to show the inductive step?

Thank you in advance!

Martin Sleziak
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Stefan Hanssen
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    Show us how far you can get with the inductive step please. – DanielV Jan 07 '15 at 19:08
  • I wanted to prove with the "n+1" – Stefan Hanssen Jan 07 '15 at 19:11
  • Are you sure you have your base step correct? Think about what the "sigma" notation means: how do you start at $1$ but plug in $0$? – daOnlyBG Jan 07 '15 at 19:13
  • yes I am sure about the basic step: $\sum_{i=0}^{n} \binom{0}{0} =\frac{0!}{0!}= 2^0=1$ – Stefan Hanssen Jan 07 '15 at 19:26
  • Each entry in Pascal's triangle gets added in to the next row twice. Therefore the sum of the numbers in the next row is exactly twice as big. – Michael Hardy Jan 07 '15 at 19:29
  • Use that $\binom{n+1}{k}=\binom{n}{k}+\binom{n+1}{k-1}$ – Vladimir Fomenko Jan 07 '15 at 19:31
  • @VolodymyrFomenko That equality is wrong: try it with $;n=4;,;;k=2;$ . – Timbuc Jan 07 '15 at 20:15
  • I know this is an inductive proof, but I want to offer some combinatorial intuition as to why this result holds. Note that $\binom{n}{k}$ is the number of ways to select $k$ ones in an $n$-digit binary string. There are $2^{n}$ such binary strings. So $\sum_{i=0}^{n} \binom{n}{i}$ covers all of the $2^{n}$ strings. – ml0105 Jan 07 '15 at 20:33

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Hints: under the agreement that $\;\binom nk=0\;$ if $\;k>n\;\;or\;\;k<0\;$ , we get

$$\begin{align}&\bullet\;\;\binom{n+1}{k}=\binom n{k-1}+\binom nk\\{}\\ &\bullet\;\;\sum_{k=0}^{n+1}\binom{n+1}k=\sum_{k=0}^{n+1}\left[\binom n{k-1}+\binom nk\right]=\sum_{k=0}^n\binom nk+\sum_{k=0}^n\binom nk=2^n+2^n=2^{n+1}\end{align}$$

Timbuc
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