For all positive integers, let P(n) be the equation 2 + 4 + 6 + … + 2n = n(n + 1

Question

For all positive integers, let P(n) be the equation
        2 + 4 + 6 + … + 2n = n(n + 1)
a. Write the equation for the base case P(1) and verify that it is true.
b. Write the inductive hypothesis P(k)
c. Write the equation for P(k+1)
d-h. Prove that P(k+1) is true
(match the statement to its reason)(statement a-h = reason A-H)

a. Base Case

b. P(k)

c. P(k + 1)

d. 2 + 4 + 6 + … + 2(k + 1)

e. 2 + 4 + 6 + … + 2k + 2(k + 1)

f. k (k + 1) + 2(k + 1)

g. (k + 1) (k + 2)

h. This is the right side of P(k + 1)

Reasons:

2 + 4 + 6 + … + 2k = k(k + 1)

by inductive hypothesis

factor out like terms

expand our sequence

P(1) 2(1) = 1(1 + 1)
2 = 1(2)
2 = 2   true

2 + 4 + 6 + … + 2(k + 1) = (k + 1) ((k + 1) + 1)
2 + 4 + 6 + … + 2(k + 1) = (k + 1)(k + 2)

therefore we’ve proven it is true

left side of P(k + 1)

2 + 4 + 6 + … + 2k = k(k + 1)

by inductive hypothesis

factor out like terms

expand our sequence

P(1) 2(1) = 1(1 + 1)
2 = 1(2)
2 = 2   true

2 + 4 + 6 + … + 2(k + 1) = (k + 1) ((k + 1) + 1)
2 + 4 + 6 + … + 2(k + 1) = (k + 1)(k + 2)

therefore we’ve proven it is true

left side of P(k + 1)

Explanation / Answer

Please let me know in case of any issue.

Match:

   a => E
   b => A
   c => F
   d => H
   e => D
   f => B
   g => c
   h => G

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