Prove that

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.

 

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