Answer to Question #187963 in Discrete Mathematics for Maricel

Question #187963

MATHEMATICAL INDUCTION AND RECURRENC


Solve the following. (10 pts each)

1. Prove P(n) = n2 (n + 1)

2. Recurrence relation an = 2n with the initial term a1 = 2.


1
Expert's answer
2021-05-07T11:39:33-0400

1. Given statement is-

P(n)=n2(n+1)P(n) = n^2 (n + 1)


at n=1,P(1)=1(1+1)=2n=1, P(1)=1(1+1)=2 , Which is even


P(1) is true.

Let us assume that P(k) is true for some positive integer k,

P(k)=k2(k+1) is even         (1)P(k)=k^2(k+1) \text{ is even } ~~~~~~~~-(1)


Now put k=k+1k=k+1

P(k+1)=(k+1)2(k+2)P(k+1)=(k+1)^2(k+2)

=(k2+1+2k)(k+2)=k3+4k2+4k+2=k3+k2+3k2+4k+2=(k^2+1+2k)(k+2)\\=k^3+4k^2+4k+2\\ =k^3+k^2+3k^2+4k+2


As k3+k2k^3+k^2 is even so,

P(k+1) is true

Hence Given statements is true for all nNn\in N .


2.an=2n,a1=2a_n=2n, a_1=2


at n=2,a2=4n=2, a_2=4

at n=3,a3=6n=3, a_3=6

at n=4,a4=8n=4, a_4=8

The Required sequence is- 2,4,6,8,....2,4,6,8,....

The required recurrence relation is

an=anan1,a_n=a_n-a_{n-1}, Where n2n\ge 2 and a1=2a_1=2


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