The product of two consecutive natural numbers is always _______.a)An even numberb)An odd numberc)A prime numberd)Divisible by 3Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Explanation:

When we multiply two consecutive natural numbers, we get a result that is always an even number. This can be proved mathematically.

Let the two consecutive natural numbers be n and (n+1).

Then, the product of these two numbers will be:

n(n+1) = n^2 + n

Now, we can see that there are two cases:

Case 1: n is an even number

In this case, we can write n as 2k where k is a natural number.

So, n^2 + n = 4k^2 + 2k = 2(2k^2 + k)

Hence, the product of two consecutive natural numbers is always an even number when n is even.

Case 2: n is an odd number

In this case, we can write n as 2k+1 where k is a natural number.

So, n^2 + n = 4k^2 + 4k + 1 + 2k + 1 = 2(2k^2 + 3k + 1) + 1

Hence, the product of two consecutive natural numbers is always an odd number when n is odd.

Conclusion:

Since the product of two consecutive natural numbers is always even when n is even and odd when n is odd, the correct answer to the given question is option 'A': An even number.

Answer

Let n be a natural number, then (n + 1) will be a consecutive natural number.
⇒ Their product = n(n + 1)
So, the product would always be an even number.
If n is odd, then n + 1 would be even, their product is even.
If n is even, then n + 1 would be odd, again their product is even.

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