Question 1 (20 points) Suppose you are proving that an integer derived in a cert

Question

Question 1 (20 points) Suppose you are proving that an integer derived in a certain way is even, and that you are at a point in the proof that you have the integer expressed as 7(4k3 +4k + D+ 1. How should you rewrite this expression to show that the integer is even? 0 28k7 + 28k +7+1 028k2 +28k +8 9 2002 +k) + 8(%2 +k 1) 2(14k2 +14k + 4) Save Question 2 (20 points) Which one of the following gives a recursive algorithm for computing an, where n is a positive integer and a is a real number? procedure power(a: real number, n: positive integer) ELI, if n ::: 1 then power(a, n) :# a else power(a, n) :#a-power(a, n) procedure power(a: real number, n: positive integer) 1 for 1 to n power a power procedure powerla: real number. n: positive integer) ipower : a for r:#: 2 to n powerm a-powern procedure powería, real number. m. positive integer) itn 1 then power(a,n):-a else Power (a, n) :# a-power(a, n-1)

Explanation / Answer

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1.

7(4k^2+4k+1) + 1
7(2k+1)^2 + 1

Now, 2k is always even
Hence, 2k+1 is always odd
Therefore, (2k+1)^2 is always odd
So, 7*(2k+1)^2 is always odd, as odd*odd multiplication is odd
So, 7*(2k+1)^2 +1 is even, as sum of 2 odd is even

BUt , we can write the expression as:
7(4k^2+4k+1) + 1
=(28k^2+28k+7) + 1
=28k^2+28k+8

Since every coefficient is divisible by 2 , it is an even number

Hence, option B is correct

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