Prove the following statement by mathematical induction, 1 n 1-2 +2²3 + ··· + n(n+1) = n+1 ... Prove that this sum has the explicit formula n That is, prove that n+1 Vn e Z+, P (n): 1 i(i+1) KONING n n+1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Prove the following statement by mathematical induction,
1
1² 2 + 2 =3 +...+7(7²+1)
Prove that this sum has the explicit formula
Vn € Z+, P (n):
n
n+1.
1
1 1
That is, prove that
i(i+1)
n+1
-
22
n+1
For full credit you must follow the form of mathematical induction, as shown in class.
You must formally begin and end your proof. You must identify the basis step, the
inductive step, and all assumptions, deductions, and conclusions. You must give your
proof line-by-line, with each line a statement with its justification.
You can use the Canvas math editor or write your math statements in English. For
example, the statement to be proved was written in the Canvas math editor. In English
it would be:
The sum of 1/(i(i+1)) for all i where 1 <=i<= n has the explicit formula n/(n+1).
Transcribed Image Text:Prove the following statement by mathematical induction, 1 1² 2 + 2 =3 +...+7(7²+1) Prove that this sum has the explicit formula Vn € Z+, P (n): n n+1. 1 1 1 That is, prove that i(i+1) n+1 - 22 n+1 For full credit you must follow the form of mathematical induction, as shown in class. You must formally begin and end your proof. You must identify the basis step, the inductive step, and all assumptions, deductions, and conclusions. You must give your proof line-by-line, with each line a statement with its justification. You can use the Canvas math editor or write your math statements in English. For example, the statement to be proved was written in the Canvas math editor. In English it would be: The sum of 1/(i(i+1)) for all i where 1 <=i<= n has the explicit formula n/(n+1).
Expert Solution
Step 1

Let P(n): 11·2+12·3+···+1n(n+1)=nn+1.

Let n=1.

Now 11(1+1)=11+1.

So P(1) is true. 

Let P(k) is true for some k1.

So  11·2+12·3+···+1k(k+1)=kk+1.

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