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Which statement about the simplified binomial expansion of [tex]$(a+b)^n$[/tex], where [tex]$n$[/tex] is a positive integer, is true?

A. The value of the binomial coefficient [tex]${ }_n C_0$[/tex] is [tex][tex]$n-1$[/tex][/tex] for all values of [tex]$n$[/tex].
B. The values of [tex]${ }_n C_1$[/tex] and [tex][tex]${ }_n C_{n-1}$[/tex][/tex] are equal to 1.
C. The values of [tex]${ }_n C_0$[/tex] and [tex]${ }_n C_n$[/tex] are equal to 1.
D. The value of the binomial coefficient [tex][tex]${ }_n C _1$[/tex][/tex] is [tex]$n-1$[/tex] for all values of [tex]$n$[/tex].

Sagot :

GinnyAnsweridn
To solve the problem, let's analyze the statements provided by examining properties of the binomial coefficients [tex]\( \binom{n}{k} \)[/tex].

### Binomial Coefficient Properties
The binomial coefficient [tex]\( \binom{n}{k} \)[/tex] is defined as:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]

We need to verify the truth of the statements based on this definition.

### Statement Analysis

1. Statement 1: "The value of the binomial coefficient [tex]\(\binom{n}{0}\)[/tex] is [tex]\(n-1\)[/tex] for all values of [tex]\(n\)[/tex]."
- Calculating [tex]\(\binom{n}{0}\)[/tex]:
[tex]\[ \binom{n}{0} = \frac{n!}{0!(n-0)!} = \frac{n!}{1 \cdot n!} = 1 \][/tex]
- This means [tex]\(\binom{n}{0}\)[/tex] is equal to 1 for any [tex]\(n\)[/tex], not [tex]\(n-1\)[/tex].
- This statement is false.

2. Statement 2: "The values of [tex]\(\binom{n}{1}\)[/tex] and [tex]\(\binom{n}{n-1}\)[/tex] are equal to 1."
- Calculating [tex]\(\binom{n}{1}\)[/tex]:
[tex]\[ \binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n!}{n} = n \][/tex]
- Calculating [tex]\(\binom{n}{n-1}\)[/tex]:
[tex]\[ \binom{n}{n-1} = \frac{n!}{(n-1)! \cdot 1!} = \frac{n!}{(n-1)! \cdot 1} = n \][/tex]
- Both [tex]\(\binom{n}{1}\)[/tex] and [tex]\(\binom{n}{n-1}\)[/tex] are equal to [tex]\(n\)[/tex], not 1.
- This statement is false.

3. Statement 3: "The values of [tex]\(\binom{n}{0}\)[/tex] and [tex]\(\binom{n}{n}\)[/tex] are equal to 1."
- Calculating [tex]\(\binom{n}{0}\)[/tex]:
[tex]\[ \binom{n}{0} = \frac{n!}{0!(n-0)!} = \frac{n!}{1 \cdot n!} = 1 \][/tex]
- Calculating [tex]\(\binom{n}{n}\)[/tex]:
[tex]\[ \binom{n}{n} = \frac{n!}{n! \cdot 0!} = \frac{n!}{n! \cdot 1} = 1 \][/tex]
- Both [tex]\(\binom{n}{0}\)[/tex] and [tex]\(\binom{n}{n}\)[/tex] are equal to 1.
- This statement is true.

4. Statement 4: "The value of the binomial coefficient [tex]\(\binom{n}{1}\)[/tex] is [tex]\(n-1\)[/tex] for all values of [tex]\(n\)[/tex]."
- Calculating [tex]\(\binom{n}{1}\)[/tex]:
[tex]\[ \binom{n}{1} = \frac{n!}{1!(n-1)!} = \frac{n!}{n} = n \][/tex]
- [tex]\(\binom{n}{1}\)[/tex] is equal to [tex]\(n\)[/tex], not [tex]\(n-1\)[/tex].
- This statement is false.

### Conclusion

After analyzing all the statements, the correct statement about the simplified binomial expansion of [tex]\((a+b)^n\)[/tex] is:

The values of [tex]\(\binom{n}{0}\)[/tex] and [tex]\(\binom{n}{n}\)[/tex] are equal to 1.
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