is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

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=2$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=2$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 $a_1=2$