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Question
Prove that `sum_(r-0)^n 3^r ""^nC_r = 4^n`
Solution
`sum_(r=0)^n 3^r ""^nC_r = 3^circ ^nC_0 + 3^1 ^nC_1 + 3^2 ^nC_2 + ..... + 3^n . ^nC_n`
= `1 + ^nC_1. 3 + ^nC_2. 3^2 + .... ^nC_n . 3^n`
= `(1 + 3)^n = 4^n`
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