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?
(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