Question Find the third term in $(x+y)^{8}$ when $x=0.3$ and $y=0.7$ .

Studdy Solution

STEP 1

1. The expression $(x+y)^{8}$ can be expanded using the binomial theorem.
2. The third term in the expansion refers to the term with $k=2$ when using the binomial theorem, since the first term corresponds to $k=0$ .
3. The values of $x$ and $y$ are given as $x=0.3$ and $y=0.7$ .
4. The binomial coefficient can be calculated using the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ .

STEP 2

1. Identify the general form of the third term in the binomial expansion of $(x+y)^{8}$ .
2. Calculate the binomial coefficient for the third term.
3. Substitute the given values of $x$ and $y$ into the third term.
4. Compute the numerical value of the third term.

STEP 3

Identify the general form of the third term in the binomial expansion of $(x+y)^{8}$ .
The third term corresponds to $k=2$ in the expansion:
$(x+y)^{8} = \sum_{k=0}^{8} \binom{8}{k} x^{8-k}y^k$
Therefore, the third term is:
$\binom{8}{2} x^{8-2}y^2$

STEP 4

Calculate the binomial coefficient for the third term.
$\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28$

STEP 5

Substitute the given values of $x$ and $y$ into the third term.
Using $x=0.3$ and $y=0.7$ , the third term becomes:
$28 \times (0.3)^{6} \times (0.7)^2$

STEP 6

Compute the powers of $x$ and $y$ .
$(0.3)^{6} = 0.3 \times 0.3 \times 0.3 \times 0.3 \times 0.3 \times 0.3$ $(0.7)^2 = 0.7 \times 0.7$

STEP 7

Calculate the numerical values of the powers of $x$ and $y$ .
$(0.3)^{6} = 0.000729$ $(0.7)^2 = 0.49$

STEP 8

Multiply the binomial coefficient by the powers of $x$ and $y$ to get the numerical value of the third term.
$28 \times 0.000729 \times 0.49$

STEP 9

Perform the multiplication to find the numerical value.
$28 \times 0.000729 \times 0.49 = 0.00994764$
The numerical value of the third term is approximately $0.00994764$ , which rounds to $0.010$ when considering the given options.
Therefore, the correct answer is A. 0.010.

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