Consider the following recurrence relation: -{+ C(n) = if n = 0 n+ 3. C(n-1) if n > 0. Prove by induction that C(n) = 3+1 -2n-3 for all n ≥ 0. 4 30+1 (Induction on n.) Let f(n) = 2n-1 4 Base Case: If n = 0, the recurrence relation says that C(0) = 0, and the formula says that f(0) = Inductive Hypothesis: Suppose as inductive hypothesis that C(k-1)=f(k-1) Inductive Step: Using the recurrence relation, C(K) = k + 3 = k + 3 X C(k-1), by the second part of the recurrence relation 3k-1+12(k-1) - 3 by inductive hypothesis X 0+1 for some k > 0. -2.0-3 , so they match.

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Need help with induction on N,  and inductive hypothesis, they are incorrect. 

Consider the following recurrence relation:
-{+
C(n) =
if n = 0
n+ 3. C(n-1) if n > 0.
Prove by induction that C(n) = 3+1 -2n-3 for all n ≥ 0.
4
30+1
(Induction on n.) Let f(n) =
Base Case: If n = 0, the recurrence relation says that C(0) = 0, and the formula says that f(0) =
Inductive Hypothesis: Suppose as inductive hypothesis that C(k-1) = f(k-1)
=k+3.
Inductive Step: Using the recurrence relation,
C(K) = k + 3. C(k-1), by the second part of the recurrence relation
= 4k +
3-1+1.
3k+1
3k+1
4
2n-1
4
4
2k-3
- 6k-3
-2(k-1)-3
4
X
so, by induction, C(n) = f(n) for all n ≥ 0.
by inductive hypothesis
X
0+1
for some k > 0.
-2.0-3
, so they match.
Transcribed Image Text:Consider the following recurrence relation: -{+ C(n) = if n = 0 n+ 3. C(n-1) if n > 0. Prove by induction that C(n) = 3+1 -2n-3 for all n ≥ 0. 4 30+1 (Induction on n.) Let f(n) = Base Case: If n = 0, the recurrence relation says that C(0) = 0, and the formula says that f(0) = Inductive Hypothesis: Suppose as inductive hypothesis that C(k-1) = f(k-1) =k+3. Inductive Step: Using the recurrence relation, C(K) = k + 3. C(k-1), by the second part of the recurrence relation = 4k + 3-1+1. 3k+1 3k+1 4 2n-1 4 4 2k-3 - 6k-3 -2(k-1)-3 4 X so, by induction, C(n) = f(n) for all n ≥ 0. by inductive hypothesis X 0+1 for some k > 0. -2.0-3 , so they match.
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