To solve the given expression:
[tex]\[ 3^{17} \binom{17}{0} - 4 \cdot 3^{16} \binom{17}{1} + 4^2 \cdot 3^{15} \binom{17}{2} - 4^3 \cdot 3^{14} \binom{17}{3} + \ldots - 4^{17} \binom{17}{17} \][/tex]
we will proceed step-by-step.
1. Identify the General Form: The expression is a combination of powers of 3 and 4, and binomial coefficients [tex]\(\binom{17}{i}\)[/tex]. It alternates in sign and follows a pattern. The general term can be represented as:
[tex]\[
(-1)^i \cdot 4^i \cdot 3^{17-i} \cdot \binom{17}{i}
\][/tex]
for [tex]\(i\)[/tex] ranging from 0 to 17.
2. Sum of the Series: The sum we want to calculate is:
[tex]\[
\sum_{i=0}^{17} (-1)^i \cdot 4^i \cdot 3^{17-i} \cdot \binom{17}{i}
\][/tex]
3. Factor Out Common Terms:
Notice that [tex]\(3^{17}\)[/tex] is a common factor across all terms when factored correctly. We can factor it out:
[tex]\[
3^{17} \sum_{i=0}^{17} \left( \frac{4}{3} \right)^i \cdot (-1)^i \cdot \binom{17}{i}
\][/tex]
4. Simplify the Series:
We recognize that this sum is related to the binomial expansion of [tex]\( (1 + x)^n \)[/tex] for [tex]\( x = -\frac{4}{3} \)[/tex]:
[tex]\[
\sum_{i=0}^{n} \binom{n}{i} x^i = (1+x)^n
\][/tex]
Here, [tex]\( x = -\frac{4}{3} \)[/tex] and [tex]\( n = 17 \)[/tex]:
[tex]\[
\sum_{i=0}^{17} \binom{17}{i} \left( -\frac{4}{3} \right)^i = \left( 1 - \frac{4}{3} \right)^{17}
\][/tex]
5. Simplify the Exponent:
Simplify the term within the parentheses:
[tex]\[
1 - \frac{4}{3} = -\frac{1}{3}
\][/tex]
Raise this to the power of 17:
[tex]\[
\left( -\frac{1}{3} \right)^{17} = -\frac{1}{3^{17}}
\][/tex]
6. Final Calculation:
Substitute this back into our factored expression:
[tex]\[
3^{17} \sum_{i=0}^{17} \left( \frac{4}{3} \right)^i \cdot (-1)^i \cdot \binom{17}{i} = 3^{17} \cdot \left( -\frac{1}{3^{17}} \right) = 3^{17} \cdot -\frac{1}{3^{17}} = -1
\][/tex]
Thus, the result of the given series is:
[tex]\[
\boxed{-1}
\][/tex]
