QUESTION 4 Drag and sequence the lines below to form a correct proof of the given claim: For every integer n 2 1, 10n-1 is divisible by9. Assume that the statement P(k is true for some k 21; that is, 1ok-1 is divisible by 9, which also means that 10k.1-9q for some integer q. 2 To examine the statement P(k+1 ), we must work with 10k+ 1-1. Algebraic calculations tell us that 10k +1-1-10 (10k-1) + gebraic calculations tell us that 101.110 (10k.1)+ 1. When n 1, we get 101.1 10-1 9, which is divisible by 9, so P(1) is true. 5. Thus, we have shown that P(k+1) holds true and completed the induction step. 4. Substitution tells us that 10k+1. 1 10(9)+9 910q+ 1), so 10k1. 1 is divisible by 9. Click Save and Submit to save and submit. Click Save All Answers to save all answers