How can I find, or what is a good reference for: $${}_nC^0 + {}_nC^1 + {}_nC^2 + \cdots + {}_nC^n = 2^n$$
I could write References [1] Binomial sums, Binomial Sums -- from Wolfram MathWorld
but I need something that looks more professional.
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Are you looking to prove this result? – Jun 30 '24 at 04:34
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3Hint: $(1+x)^n = {}^nC_0+{}^nC_1 x + {}^nC_2x^2 \cdots {}^nC_n x^n$ Also i think u mean $2^n$ – RandomGuy Jun 30 '24 at 04:35
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2This question is similar to: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – peterwhy Jun 30 '24 at 04:39
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4Why do you need a reference that « looks more professional » ? This is a very well known result that appears in any undergraduate text. If you want to cite a book you can take Serge Lang - Basic Mathematics for example. – Gribouillis Jun 30 '24 at 06:02
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1DLMF equation https://dlmf.nist.gov/1.2.E3 professional enough? – Somos Jun 30 '24 at 16:17
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Yes. That sounds good. Thank you for help. – Mocean Jul 02 '24 at 03:54
2 Answers
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As @randomguy had said in the comments, you can use the Binomial expansion of $(1+x)^n$
If you cannot find it, here's the answer
Substituting $x=1$ in the hint, it yields
$(1+1)^n=C^n_0+C^n_1+C^n_2+....C^n_n $
$\Rightarrow 2^n=C^n_0+C^n_1+C^n_2+....C^n_n$
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As we know from Binomial Theorm that,
$(1+x)^n={}^nC_0 \ x+{}^nC_1 \ x^2+{}^nC_2 \ x^3+\cdots+{}^nC_n \ x^n$
$\therefore (1+1)^n={}^nC_0 \ (1)+{}^nC_1 \ (1)^2+{}^nC_2 \ (1)^3+\cdots+{}^nC_n \ (1)^n$
$\Rightarrow 2^n={}^nC_0+{}^nC_1+{}^nC_2+\cdots+{}^nC_n$
Fetray
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