Show that [r=0,n]ΣnCr 2^r=3^n
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Show that [r=0,n]ΣnCr 2^r=3^n

[From: ] [author: ] [Date: 11-12-16] [Hit: ]
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So I am thinking :

(a+b)^n=[r=0, n]Σ nCr a^n-1*b^r

Since 'a' is not given explicitly I take it is a=1 and b=2 in [r=0,n]ΣnCr 2^r=3^n therefore (1+2)^n=3^n

Do you think this way of showing that [r=0,n]ΣnCr 2^r=3^n would work?

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Hint: Let a = 1, b = 2
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keywords: Sigma,that,Show,nCr,Show that [r=0,n]ΣnCr 2^r=3^n
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