Asked by Iris
Which of the following shows the correct first step to prove the following by mathematical induction?
3 + 11 + 19 + 27 + … + (8n - 5) = n(4n - 1)
A. 3 + 11 + 19 + 27 + … + (8 • 1 - 5) = 1(4 • 1 - 1)
B. 8 • 1 - 5 = 1(4 • 1 - 1)
C. 3 + 11 + 19 + 27 + … + (8k - 5) = k(4k - 1)
D. 3 + 11 + 19 + 27 + … + (8k - 5) + [8(k + 1) - 5] = (k + 1)[4(k + 1) - 1]
Thank you
3 + 11 + 19 + 27 + … + (8n - 5) = n(4n - 1)
A. 3 + 11 + 19 + 27 + … + (8 • 1 - 5) = 1(4 • 1 - 1)
B. 8 • 1 - 5 = 1(4 • 1 - 1)
C. 3 + 11 + 19 + 27 + … + (8k - 5) = k(4k - 1)
D. 3 + 11 + 19 + 27 + … + (8k - 5) + [8(k + 1) - 5] = (k + 1)[4(k + 1) - 1]
Thank you
Answers
Answered by
Reiny
I don't know what procedure you learned, but this is how I taught this:
step 1:
show it to be true for n = 1
LS = 3
RS = 1(4-1) = 3
step 2:
assume it is true for n = k
that is
3+11+19+..+ (8k-5) = k(4k - 1)
step 3:
show it then must be true for n = k+1
that is, show
3+11+19+ ... + (8k-5) + (8(k+1) - 5) = (k+1)(4(k+1) - 1)
or
3+11+19+ ... + (8k-5) + (8k+3) = (k+1)(4k+3)
LS = 3+11+19+ ... + (8k-5) + (8k+3)
= k(4k-1) + 8k+3
= 4k^2 - k + 8k + 3
= 4k^2 + 7k + 3
RS = (k+1)(4k+3)
= 4k^2 + 3k + 4k+ 3
= 4k^2 + 7k + 3
step 1:
show it to be true for n = 1
LS = 3
RS = 1(4-1) = 3
step 2:
assume it is true for n = k
that is
3+11+19+..+ (8k-5) = k(4k - 1)
step 3:
show it then must be true for n = k+1
that is, show
3+11+19+ ... + (8k-5) + (8(k+1) - 5) = (k+1)(4(k+1) - 1)
or
3+11+19+ ... + (8k-5) + (8k+3) = (k+1)(4k+3)
LS = 3+11+19+ ... + (8k-5) + (8k+3)
= k(4k-1) + 8k+3
= 4k^2 - k + 8k + 3
= 4k^2 + 7k + 3
RS = (k+1)(4k+3)
= 4k^2 + 3k + 4k+ 3
= 4k^2 + 7k + 3
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