Download MATH 1390 W12 Quiz 4 Solutions 1 1. Using mathematical

MATH 1390 W12 Quiz 4 Solutions
1
1. Using mathematical induction, show that for all integers n ≥ 1
1 + 7 + 13 + 19 + · · · + (6n − 5) = n(3n − 2).
Let P (n) be the statement: 1 + 7 + 13 + 19 + · · · + (6n − 5) = n(3n − 2).
a State what needs to be proven in the base step.
Solution We need to show that P (1) is true; that is, the above equation holds for
n = 1.
b) Prove the base step.
Solution When n = 1, the left side (L.S.) of the equation equals 1. The right side
(R.S.) equals 1 · (3 · 1 − 2) = 1. ∴ L.S. = R.S. and P (1) is true.
c) State what needs to be proven in the inductive step.
Solution We need to show for all n ≥ 1 if P (n) is true, then P (n + 1) is true.
That is, if the equation holds for n then it holds for n + 1.
d) Prove the inductive step.
Suppose the equation holds for some n ≥ 1. Then 1 + 7 + 13 + 19 + · · · + (6n − 5) =
n(3n − 2). We need to show that the equation holds for n + 1; that is,
1 + 7 + 13 + 19 + · · · + (6(n + 1) − 5) = (n + 1)(3(n + 1) − 2).
The L.S. is 1+7+13+19+· · ·+(6n−5)+(6(n+1)−5) = n(3n−2)+6(n+1)−5. Now
n(3n − 2) + 6(n + 1) − 5 = 3n2 + 4n + 1. But on the R.S. (n + 1)(3(n + 1) − 2) =
3n2 + 4n + 1. So L.S. = R.S. and hence P (n + 1) is true. This completes the
induction.
2. Using mathematical induction, show that 6n − 1 is divisible by 5.
Let P (n) be the statement: 5 divides 6n − 1.
a State what needs to be proven in the base step.
Solution Need to show that P (0) is true.
b) Prove the base step.
Solution 60 − 1 = 1 − 1 = 0. Since 50, this shows that P (0) is true.
c) State what needs to be proven in the inductive step.
for all n ≥ 0, if P (n) is true, then P (n + 1) is true.
Solution Need
n to show that
n+1
That is, if 5 6 − 1, then 5 6
− 1.
d) Prove the inductive step.
Solution Suppose 5 divides 6n − 1 for some n ≥ 0. We observe
that 6n+1 − 1 =
6· 6n − 1 = (5 + 1)6n − 1 = 56n + 6n − 1. Since 556n and 56n − 1 it follows that
56n+1 − 1.
Bonus Show that 2n < (n + 1)! for all integers n ≥ 2.
MATH 1390 W12 Quiz 4 Solutions
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Solution By induction. The assertion is seen to be true for n = 2 since 22 = 4 < 3! = 6.
Suppose 2n < (n + 1)! for some n ≥ 2. Then since n + 2 ≥ 4 we have that
2n+1 < 2(n + 1)! < (n + 1)!(n + 2) = (n + 2)!.
So 2n+1 < (n + 2)! and the proof follows by induction.