Answer:
P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Step-by-step explanation:
Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3
Basis: P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:
2(k+1) - 1 = 2k + 2 - 1
≤ 2 + k! (by the inductive hypothesis)
= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.
I think 121
406 + 94 = 500 and 180 + 199 = 379.
Subtract that and 121 is what’s left.
-29 into a fraction = -29/1.
to make a whole number into a fraction, just put it over 1
