Prove by induction the product of n odd integers is odd for n > 1. recall a number k is odd if there exists an integer m such that k= 2m+ 1.

Step-by-step explanation: Induction:

Suppose that for n odd numbers is true that their product is odd.

Lets find for n+1 numbers:

The pruduct of n+1 odd numbers can be written as:

(2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1)

where the numbers "k" can be any integer number.

Because of the hipotesis, we know that:

(2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1) is an odd number, so we can write this as:

(2k + 1)

Then we have: (2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) = (2k + 1)*(2kₙ₊₁ + 1)

now we have to solve this product:

(2k + 1)*(2kₙ₊₁ + 1) = 2k*2kₙ₊₁ + 2k + 2kₙ₊₁ + 1 = 2*(2k*kₙ₊₁ + k + kₙ₊₁) + 1

Because k and kₙ₊₁ are integers, the number (2k*kₙ₊₁ + k + kₙ₊₁) is also an integer, that we can write as L.

So we have that:

(2k₁ + 1)*(2k₂ +1)*...*(2kₙ + 1)*(2kₙ₊₁ + 1) = 2L + 1

So the product of n + 1 odd integers is also an odd number.  

Now you can test that this is true for n = 2 (the same procedure that I did in the product of (2k + 1)*(2kₙ₊₁ + 1) ) and there you have your demonstration by induction.

x=12 units bd= 10 units

step-by-step explanation:

answer: approximately 206 bones

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