Question:
Prove that
$\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 3^{r}=4^{n}$
Solution:
To prove $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 3^{r}=4^{n}$
Formula used: $\sum_{r=0}^{n}{ }^{n} C_{r} \cdot a^{n-r} b^{r}=(a+b)^{n}$
Proof: In the above formula if we put a = 1 and b = 3, then we will ge
$\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 1^{n-r} 3^{r}=(1+3)^{n}$
Therefore,
$\sum_{r=0}^{n}{ }^{n} C_{r} \cdot 3^{r}=(4)^{n}$
Hence Proved.