I. Use mathematical induction, a basis proof, a inductive hypothesis and step to solve:

Question

∀n ∈ N, ∑n i=1 i!(i² + 1) = (n + 1)!n.

II. Give a recursive definition for the set of all strings of 0's and 1's for which all the 0s follow all the 1s.

Answer 1

The summation formula is unclear and FALSE

For n=1,2,3...

1!*(1^2+1)=2 while (1+1)!*1=2

1!*(1^2+1)+2!(2^2+1)=2+10=12

While (2+1)!*2 = 6*2=12

1!*(1^2+1)+2!*(2^2+1) +3!*(3^2+1)=2+2*5+6*10=2+10+60=72

While

(3+1)!*3=24*3=72

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Iinduction hypothesis says it is given that for some positive integer l, the following statement holds:

P(1)+p(2)+....p(k-1) + k!*(k^2+1) = (k+1)!*k

P(1)+p(2)+.... + k!*(k^2+1) + (k+1)! *((k+1)^2+1)=

(k+1)!*k + (k+1)! *((k+1)^2+1)=

(K+1)! [ k + k^2+2k+2]

(K+1)! [ k^2 + 3k +2]

(K+1)! (K+1)(k+2)

(K+2)! (K+1)

So the statement holds for k+1

[End of proof]

2) recursive set

10 is in the set

X10y is in the set if x is 1 and y is 0

Any bit string, S , S is in the set if S' is in the set

Where S = xS' y