Question
Prove that ${\sum\limits_{r=0}^n3^r\;^nC_r=4^n}$

Answer

$\sum_{r=0}^n\;^nC_ra^{n-r}b^r=\;\left(a\;+b\right)^n\;.........\left(1\right)$
$Now,\sum_{r=0}^n3^r\;^nC_r\;=\overset n{\underset{r=0}{\sum\nolimits}}\;^nC_r\;(1)^{n-r}.3^r\;=\;(1+3)^n\;\;\;\;(by\;(1)$
$\;\;\;=\;4^n$

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