Solution to MATHEMATICAL INDUCTION AND RECURRENC Solve the following. (10 pts each) 1. Prove P(n) = n2 … - Sikademy
Author Image

Archangel Macsika

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.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

1. Given statement is-

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


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


Now put k=k+1

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

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


As k^3+k^2 is even so,

P(k+1) is true

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


2.a_n=2n, a_1=2


at n=2, a_2=4

at n=3, a_3=6

at n=4, a_4=8

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

The required recurrence relation is

a_n=a_n-a_{n-1}, Where n\ge 2 and a_1=21. Given statement is-

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


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


Now put k=k+1

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

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


As k^3+k^2 is even so,

P(k+1) is true

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


2.a_n=2n, a_1=2


at n=2, a_2=4

at n=3, a_3=6

at n=4, a_4=8

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

The required recurrence relation is

a_n=a_n-a_{n-1}, Where n\ge 2 and a_1=21. Given statement is-

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


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


Now put k=k+1

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

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


As k^3+k^2 is even so,

P(k+1) is true

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


2.a_n=2n, a_1=2


at n=2, a_2=4

at n=3, a_3=6

at n=4, a_4=8

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

The required recurrence relation is

a_n=a_n-a_{n-1}, Where n\ge 2 and a_1=2

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-3043-qpid-1742