Question

\(\forall n \in N,\left(3+5^{(1 / 2)}\right)^{n}+\left(3-5^{(1 / 2)}\right)^{\mathrm{n}}\) is (a) Even natural number (b) Odd natural number (c) Any natural number (d) Rational number

Short answer

The expression \(\left(3+5^{(1/2)}\right)^n+\left(3-5^{(1/2)}\right)^n\) simplifies to \(\sum_{m=0}^{\lfloor n/2 \rfloor} 2\binom{n}{2m} 3^{n-2m} 5^{m}\), which is always an even natural number. Therefore, the correct answer is (a) Even natural number.

Step-by-step solution

Expanding using the binomial theorem

: Applying the binomial theorem to the given expression, we get: \((3+5^{(1/2)})^n+(3-5^{(1/2)})^n = \sum_{k=0}^n \binom{n}{k} 3^{n-k} (5^{(1/2)})^k + \sum_{k=0}^n \binom{n}{k} 3^{n-k} (-1)^k (5^{(1/2)})^k\)

Combining the sums

: We can simplify this by combining the sums: \[= \sum_{k=0}^n \binom{n}{k} 3^{n-k} (5^{(1/2)})^k + \binom{n}{k} 3^{n-k} (-1)^k (5^{(1/2)})^k\] \[= \sum_{k=0}^n \binom{n}{k} 3^{n-k} (5^{(1/2)})^k (1+(-1)^k)\] Notice when \(k\) is even, the value of \((-1)^k\) is 1, and when \(k\) is odd, its value is -1. Thus, the \((1+(-1)^k)\) term will be 0 for odd \(k\) and 2 for even \(k\). Therefore, only even \(k\) values will contribute to the sum.

Simplifying the sum

: As only even \(k\) contribute, we can rewrite the sum using \(k=2m\), where \(m\) is an integer: = \(\sum_{m=0}^{\lfloor n/2 \rfloor} 2\binom{n}{2m} 3^{n-2m} (5^{(1/2)})^{2m}\) Now we can simplify the expression: = \(\sum_{m=0}^{\lfloor n/2 \rfloor} 2\binom{n}{2m} 3^{n-2m} 5^{m}\)

Analyzing the expression

: Now let's analyze the simplified expression for evenness, oddness, and rationality. - Evenness: The expression will always be even, as each term in the sum contains a factor of 2. So, the expression is an even natural number. - Rationality: All terms in the sum are integers since each term is a product of an integer binomial coefficient and positive integer powers of 3 and 5. Therefore, the resulting sum is also an integer, and thus a rational number. So, the correct answer is (a) Even natural number.

Key concepts

Even Natural Number

An even natural number is any number that belongs to the set of natural numbers and is divisible by 2 without a remainder. In simpler terms, these numbers can be expressed as double another integer. For example, 2, 4, 6, and 8 are even natural numbers because dividing them by 2 leaves no remainder.

In mathematics, recognizing whether a number is even or odd helps in solving various problems, as it impacts the outcome of calculations such as addition, subtraction, and multiplications.

In the context of the exercise provided, the expression is always even because it includes a factor of 2 in each term due to the sum containing the coefficient 2. This factor ensures divisibility by 2 for every component in the sequence, making the final result an even natural number.

Rational Numbers

Rational numbers are numbers that can be expressed as the quotient \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q eq 0 \). Hence, rational numbers include integers, fractions, and finite decimals.

One key property of rational numbers is that they can be both positive and negative. They form a dense set, meaning between any two rational numbers, there exists another rational number.

In our problem, the expanded expression of the binomial involves terms that are inherently integers (combinations of products of integers). Therefore, the sum is an integer, and since integers are special types of rational numbers (where \( q=1 \)), the expression also falls into the category of rational numbers.

Integer Powers

Integer powers represent expressing a number as being multiplied by itself a certain number of times. For instance, \(3^2\) means 3 is multiplied by itself (3 \times 3) to give 9. Including negative exponents means performing repeated division instead.

Integer powers are vital in calculations involving growth, scaling, and other mathematical modeling. They help in simplifying and analyzing expressions like in the binomial theorem.

In the exercise solution, integer powers of 3 and 5 appear in the terms \(3^{n-k}\) and \(5^{m}\). Since both the base numbers (3 and 5) are integers and the exponents are non-negative integers, each calculation results in another integer. This characteristic contributes to the expression's rationality and determines the outcome as a natural number.