MATH220 April 2011
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q2 • Q3 (a) • Q3 (b) • Q4 • Q5 • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q8 • Q9 (a) • Q9 (b) • Q9 (c) • Q10 (a) • Q10 (b) •
Question 07 (a)
Let be a sequence of real numbers defined by
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We will prove this by induction.
The first case, is easy to go through, indeed
so the first term of the sequence matches the general formula. You can do as many initial cases as you want, but remember, you have to at least one to get your argument started.
Then comes the general case. We say: assume that the statement is true for all values of up to . We will show that the statement is also true for .
By definition of the sequence , we know that
And we also assume that the statement is true up to hence
Combining these two facts together, we obtain that
which is the claimed statement to be proved and so concludes our proof.
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