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Greatest Binomial Coefficients

Prove that ∑ ...

Question

Prove that $\sum_{r = 0}^{n} 3^{r} n_{C_{r}} = 4^{n}$ .

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Solution

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

$3^{0} n_{C_{0}} + 3^{1} n_{C_{1}} + 3^{2} n_{C_{2}} + . . . + 3^{n} n_{C_{n}}$

$= n_{C_{0}} (1)^{n} . 3^{0} + n_{C_{1}} (1)^{n - 1} {.3}^{1} + n_{C_{2}} (1)^{n - 2} . 3^{2} + . . . + n_{C_{n}} 3^{n}$

$= (1 + 3)^{n} = 4^{n}$

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n \sum r = 0 n_{C_{r}} 3^{r} = 4^{n}

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n \sum r = 0 3^{r}^{n} C_{r} = 4^{n}

n \sum r = 0 3^{r}

^{n} C_{r} = 4^{n}

^{4 n} C_{2 n} :^{2 n} C_{n} = 1.3.5... (4 n - 1) : [1.3.5... (2 n - 1)]^{2} .

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Greatest Binomial Coefficients

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