Question

the sum of the squares of three consecutive odd numbers increased by 1 is divisible by(use mathematical induction)
(1) 12 as well as 24
(2) 12 but not 24
(3)neither 12 nor 24
(4)by all multiples of 12​

Answer 1

Answer:

12 but not 24 is correct option

Step-by-step explanation:

the sum of the squares of three consecutive odd numbers increased by 1 is divisible by

(2n - 1) , (2n + -) , (2n + -)

(2n - 1)² + (2n + 1)² + (2n + 3)²

= 4n² + 1 - 4n + 4n² + 1 + 4n + 4n² + 9 + 12n

= 12n² + 12n + 11

if sum is increased by 1

then

= 12n² + 12n + 11 + 1

= 12n² + 12n + 12

= 12(n² + n + 1)

it is divisible by 12

to be divisible by 24 , n² + n + 1 should be even

n² + n + 1 = (n + 1)² - n

if n is even then  n+1 is odd  square of odd is odd

odd - even = odd

if n is odd then n + 1 is even square of even is even

even - odd = odd

=> n² + n + 1 can not be even

so not divisible by 24

12 but not 24 is correct option