We are presenting a consolidated answer here:
Let the given statement be P(n). Then we need to prove that:
P(k): (sn - tn) is divisible by (s -
t).
When n = 1, the given statement becomes: (s1 -
t1) is divisible by (s - t), which stands true, hence
P(1) is true.
Let p(k) be true. Then P(k): sk -
tk is divisible by (s-t).
Now, sk + 1 - tk + 1 = sk + 1 -
skt + skt - tk + 1 [on adding and
subtracting skt]
= sk(s - t) + t(sk - tk), which is
divisible by (s - t) [using (i)]
⇒ P(k + 1): sk + 1 - tk + 1is divisible by (s
- t)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus as P(1) and P(k + 1) stand true, whenever P(k) is true. Hence,
by the Principal of Mathematical Induction, P(n) is true for all n
∈ N.
5. 14 marks In this problem we will explore the factors of expression a. 12 marks] Suppose a fr...
5. 「14 marks! In this problem we will explore the factors of expression sr_ t C. 12 marks| Suppose your friend now claims "(s+t) divides sk th, when k is even." Write this claim as a theorem, specifying the domains for each of the variables and using the appropriate quantifiers d. 15 marks Either prove or disprove the theorem from the previous part
5. 「14 marks! In this problem we will explore the factors of expression sr_ t
C. 12...
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