Calculate the value of (0.8)4 using the binomial theorem.

Question

ScanSolve AI Tutor · Accepted Answer

Let's solve this problem again, Ferdinand.

To calculate the value of $(0.8)^4$ using the binomial theorem, we first express 0.8 as a difference from 1.

Rewrite the expression in the form $(a+b)^n$ .
We can write 0.8 as $1 - 0.2$ . So, we need to expand $(1 - 0.2)^4$ .
Here, $a=1$ , $b=-0.2$ , and $n=4$ .
Apply the binomial theorem formula.
The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ .
For $(1 - 0.2)^4$ , the expansion is:
$(1 - 0.2)^4 = \binom{4}{0} (1)^4 (-0.2)^0 + \binom{4}{1} (1)^3 (-0.2)^1 + \binom{4}{2} (1)^2 (-0.2)^2 + \binom{4}{3} (1)^1 (-0.2)^3 + \binom{4}{4} (1)^0 (-0.2)^4$
Calculate the binomial coefficients.
$\binom{4}{0} = 1$ $\binom{4}{1} = 4$ $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ $\binom{4}{3} = 4$ $\binom{4}{4} = 1$
Substitute the coefficients and calculate each term.
$1 \cdot (1)^4 \cdot (-0.2)^0 = 1 \cdot 1 \cdot 1 = 1$ $4 \cdot (1)^3 \cdot (-0.2)^1 = 4 \cdot 1 \cdot (-0.2) = -0.8$ $6 \cdot (1)^2 \cdot (-0.2)^2 = 6 \cdot 1 \cdot (0.04) = 0.24$ $4 \cdot (1)^1 \cdot (-0.2)^3 = 4 \cdot 1 \cdot (-0.008) = -0.032$ $1 \cdot (1)^0 \cdot (-0.2)^4 = 1 \cdot 1 \cdot (0.0016) = 0.0016$
Sum all the terms.
$(1 - 0.2)^4 = 1 - 0.8 + 0.24 - 0.032 + 0.0016$ $(1 - 0.2)^4 = 0.2 + 0.24 - 0.032 + 0.0016$ $(1 - 0.2)^4 = 0.44 - 0.032 + 0.0016$ $(1 - 0.2)^4 = 0.408 + 0.0016$ $(1 - 0.2)^4 = \boxed{0.4096}$

Answer