| Let P(n) be the claim: | n | (2i - 1) | = |
n2 | for all integers n 1. |
| S | |||||
| i = 1 | |||||
2a)
| Let P(n) be the claim: | n | (2i - 1) | = |
n2 | for all integers n1. |
| S | |||||
| i = 1 | |||||
| P(1) is the statement: | 1 | (2i - 1) | = |
12. |
| S | ||||
| i = 1 | ||||
| P(k) is the statement: | k | (2i - 1) | = |
k2. |
| S | ||||
| i = 1 | ||||
| P(k+1) is the statement: | k + 1 | (2i - 1) | = |
(k + 1)2. |
| S | ||||
| i = 1 |
For a proof by induction, you first need to check that the statement P(1) is true. Then assume that P(k) is true and use this to show that P(k + 1) is true.
| The statement P(1) is true since | 1 | (2i - 1) | = |
(2·1 - 1) = 1 = 12. |
| S | ||||
| i = 1 |
Now assume that the statement P(k) is true. We now need to show that the left-hand side of P(k+1) is equal to the right-hand side of P(k+1).
| L.H.S of P(k+1) | = |
k + 1 | (2i - 1) | ||
| S | |||||
| i = 1 | |||||
= |
k | (2i - 1) | + |
[2 (k + 1) - 1] | |
| S | |||||
| i = 1 | |||||
| = | k2 + [2 (k + 1) - 1] (since we are assuming that P(k) is true) | ||||
| = | k2 + 2k - 1 | ||||
| = | (k + 1)2 | ||||
| = | R.H.S. of P(k+1) | ||||
| Therefore, for all integers n 1, |
n | (2i - 1) | = |
n2. |
| S | ||||
| i = 1 |