To evaluate the sum [tex]\(\sum_{k=0}^{2001}\binom{2001}{k} 3^k(-4)^{2001-k}\)[/tex], observe the following steps:
1. Identify the Binomial Theorem Application:
The given sum can be identified as an expression derived from the binomial theorem. The binomial theorem states that for any integers [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and any non-negative integer [tex]\(n\)[/tex]:
[tex]\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}\][/tex]
2. Apply the Binomial Theorem:
In our case, we have [tex]\(a = 3\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(n = 2001\)[/tex]. By substituting these values into the binomial theorem, we get:
[tex]\[
(3 + (-4))^{2001} = \sum_{k=0}^{2001} \binom{2001}{k} 3^k (-4)^{2001-k}
\][/tex]
3. Simplify the Expression:
Now simplify the expression inside the parentheses:
[tex]\[
3 + (-4) = -1
\][/tex]
Thus, the expression becomes:
[tex]\[
(-1)^{2001}
\][/tex]
4. Evaluate the Power:
Since [tex]\((-1)\)[/tex] raised to any integer power is either [tex]\(-1\)[/tex] or [tex]\(1\)[/tex] depending on whether the exponent is odd or even:
[tex]\[
(-1)^{2001}
\][/tex]
Here, since [tex]\(2001\)[/tex] is an odd number, [tex]\((-1)^{2001} = -1\)[/tex].
Therefore, the value of [tex]\(\sum_{k=0}^{2001}\binom{2001}{k} 3^k(-4)^{2001-k}\)[/tex] is [tex]\(-1\)[/tex].
