# Prove that 2.7n + 3.5n-5 is divisible by 24, for all n belongs to N....Plzzz dnt tell to refer textbook as frm dat also its not clear to me plzzzzzz answer it as soon as possible.....

Let p (n) be the statement given by p (1) is true

Let p (m) : 2.7m + 3.5m – 5 is divisible by 24 be true

⇒ 2.7m + 3.5m – 5 = 24λ  ;λ∈N

⇒ 3.5m = 24λ + 5 – 2.7m  ...  (1)

Now

p (m + 1) : 2.7m + 1 + 3.5m + 1 – 5

= 2.7m + 1 + (3.5m) 5  – 5

= 2.7m + 1 + (24λ + 5 – 2.7m) 5  – 5    (from (1))

= 2.7m + 1 + 120λ + 25 – 10.7m  – 5

= (2.7m + 1 – 10.7m) + 120λ + 20

= (2 × 7λ 7m – 10 × 7m) + 120λ + 24 – 4

= (14 – 10)7m – 4 + 24 (5λ + 1)

= 4 (7m – 1) + 24 (5λ + 1)

= 4 × 6µ + 24 (5λ + 1)  (∵ 7m – 1 is a multiple of 6 for all m∈N ∴ 7m – 1 = 6µ, µ∈N)

= 24 (µ +  + 1)  which is divisible by 24

p (m + 1) is true.

Hence by principle of mathematical induction p (n) is true for all n∈N

• -2

2.7n + 3.5n-5 is divisible by 24---- to prove

put n=1

14+15-4=24

p(1) true.

put n=k

2.7k + 3.5k-5=24m-------1)

to prove: n=k+1 is true

put n= k+1

2.7.7k+3.5.5k-5

from 1) -5=24m-2.7k-3.5k

14.7k+15.5k+24m-2.7k-3.5k

24m+7k(14-2)+5k(15-03)

24m+12.7k+12.5k

24(m+7k/2+5k/2)

24 multiplied by any digit means it is divisible by 24.

by p.m.i. it is proved that n is true for N

• 11
What are you looking for?