Proove that r=0 submission $r = n 3^{r} n C r = 4^{n}$

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Solution

To prove $\sum_{r = 0}^{n} 3^{r} n C_{r} = 4^{n}$

By Binomial theorem,

$\sum_{r = 0}^{n} n C_{r} a^{n - r} b^{r} = (a + b)^{n}$
From given, we take $b = 3, a = 1$

$\sum_{r = 0}^{n} 3^{r} n C_{r} (1)^{n - r} = (1 + 3)^{n}$

$\sum_{r = 0}^{n} 3^{r} n C_{r} (1) = 4^{n}$

$∴ \sum_{r = 0}^{n} 3^{r} n C_{r} = 4^{n}$

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