Exercise on Mathematical Induction Timothy Rolfe
Solution
DUE: Tuesday, 16 January 2007
Suggested by Ian Parberry and William Gasarch, Problems in Algorithms; Second Edition
(2002), p. 8 (Section 2.1, Problem 4 — revised):
Prove by induction: S(n) =
n
j
nnn
jj
1
3
21
1
This can be represented as the following recurrence:
For n = 0, S(n) = 0
For n > 0, S(n) = S(n–1) + n ( n + 1 )
Note that
3
11
)1(
nnn
nS
(1) Basis: S(0) = 0 · 1 ·2 / 6 = 0 / 6 = 0
(2) Induction:
Hint: You may find it convenient to drive your induction by assuming the formula for S(n-1),
and then proving S(n).
)(nS
1)1( nnnS
Recurrence
1
3
3
3
11
nn
nnn
Substitute the inductive
hypothesis
31
3
)1(
n
nn
Factor out
3
1nn
3
21
nnn
Simplify — and QED
Printed 2020-12 月-05 at 21:29