Question

n(n+1)(n+5) is a multiple of 3 prove by PMI​

Answer 1

Answer:

Proved!!

Step-by-step explanation:

Given Problem:

n(n+1)(n+5) is a multiple of 3 prove by PMI​.

Solution:

PMI stands for Principal of mathematical induction.

Now,

Method:

$\sf => p(n)=n(n+1)(n+5)$

$\sf =>p(1)=1(1+1)(1+5)=1x2x6=12$

Assume that p(k) b true,

$\sf =>p(k)=k(k+1)(k+5)$

On solving,

$\sf =>k3+6k2+5k=3y\quad(as\; its\; a\;multiple\; of\; 3)$

$\sf =>k3=3y-6k2-5k ---(1)$

$=>\sf p(k+1)= k+1 ( k+1 +1) k+1 +5)$

$\sf =>k+1 ( k+2) (k+6)$

On solving,

$\sf =>k3+9k2+20k +12$

Putting value of k3 from (1),

$\sf =>3y-6k2-5k+9k2+20k +12$

$\sf =>3y+3k2+15k +12$

$\sf =>3(y+k25k +4)$

Therefore,

We can say p(k+1) is a multiple of 3

p(k) is true so p(n) is also true and its a multiple of 3 p(n)=n(n+1)(n+5)

p(1)=1(1+1)(1+5)=1x2x6=12

Let p(k) b true,

p(k)=k(k+1)(k+5)

On solving,

$\sf =>k3+6k2+5k=3y\quad(As\; its\; a\; multiple\;of\;3)\\\ =>k3=3y-6k2-5k ------(1)\\\ =>p(k+1)= k+1 ( k+1 +1) k+1 +5)\\\ =>k+1 ( k+2) k+6)$

On solving,

$\sf k3+9k2+20k +12$

Now,

Putting value of k3 from (1),

$\sf =>3y-6k2-5k+9k2+20k +12\\\ =>3y+3k2+15k +12\\\ =>3(y+k25k +4)$

Therefore,

We can say p(k+1) is a multiple of 3.

p(k) is true so p(n) is also true and its a multiple of 3

Hence Proved!!