Algebra

[hushed bobcat]

How to find the term if the problem is in fraction?

[spring schooner]

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[cursive hawk]

that's an unfortunate wording

you can read the binomial expansion in two directions

[hushed bobcat]

what do you mean

[cursive hawk]

1. and 3. are ambiguous is what I mean

[hushed bobcat]

right I'm confused 😅

[cursive hawk]

$$ (a+b)^n = \sum _{k=0}^n \binom{n}{k}a^kb^{n-k} = \sum _{k=0}^n \binom{n}{n-k}a^{n-k}b^k $$

[wet rover]

aL

[cursive hawk]

answering 2. is easy at least

[hushed bobcat]

I have an answer for 1 but not sure if it is correct

[cursive hawk]

sure, what you get?

[cursive hawk]

ok this is 7 choose 3

so you are following this format

do you see the problem now?

[hushed bobcat]

no sorry

[cursive hawk]

you can switch these powers

3 and 4 instead

the coefficient is still the same

and you get 4th term in this format

[hushed bobcat]

I just follow this formula? But change (1/a²)³ and (x/2)⁴?

[cursive hawk]

yes, to get 4th term you put k = 3

that's why I say unfortunate wording

because both are correct

$$ \left( \frac{1}{a^2}+\frac{x}{2} \right)^7 = \left( \frac{x}{2} + \frac{1}{a^2} \right)^7 $$

[wet rover]

aL

[cursive hawk]

you can look at it in both ways

[hushed bobcat]

my final answer will change too right instead of 35x³/8a⁸

[cursive hawk]

yes, depending on which way you look at it

but that's really not your fault

if the teacher has a problem with it, then that's his or her problem

[hushed bobcat]

I see 😂😭

[cursive hawk]

but you can answer 2. precisely