With $n$ a positive integer, evaluate the sum
$\binom{n}{0}-3\binom{n}{1}+3^2\binom{n}{2}+\cdots+(-1)^n3^n\displaystyle\binom{n}{n}=\sum_{k=0}^n(-3)^k\binom{n}{k}$
Anyone know how to approach this problem?
With $n$ a positive integer, evaluate the sum
$\binom{n}{0}-3\binom{n}{1}+3^2\binom{n}{2}+\cdots+(-1)^n3^n\displaystyle\binom{n}{n}=\sum_{k=0}^n(-3)^k\binom{n}{k}$
Anyone know how to approach this problem?
From the definition of binomial coefficients, $\sum_{k=0}^n {n\choose k}x^k=(1+x)^n.$ For your problem, take $x=-3$ to conclude the sum is $(-2)^n$.
Recall that $(1-x)^n = \sum_{k=0}^n \dbinom{n}k (-x)^k$