Prove that $\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$.
(N/A) By the Binomial Theorem,we have the expansion:
$\sum\limits_{r = 0}^n {^nC_r a^{n - r} b^r} = (a + b)^n$
By substituting $a = 1$ and $b = 3$ in the above equation,we obtain:
$\sum\limits_{r = 0}^n {^nC_r (1)^{n - r} (3)^r} = (1 + 3)^n$
Since $(1)^{n - r} = 1$,the expression simplifies to:
$\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$
Hence,the identity is proved.
$\sum\limits_{r = 0}^n {^nC_r a^{n - r} b^r} = (a + b)^n$
By substituting $a = 1$ and $b = 3$ in the above equation,we obtain:
$\sum\limits_{r = 0}^n {^nC_r (1)^{n - r} (3)^r} = (1 + 3)^n$
Since $(1)^{n - r} = 1$,the expression simplifies to:
$\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$
Hence,the identity is proved.
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