Question

Prove by mathematical induction that the sum of cubes of three consecutive integers is divisible by 9

Answer 1

Answer:

Let n , n+1 , n+2 be three consecutive integers

Let P(n)=n³+(n+1)³+(n+2)³ is divisible by 9.

Step-by-step explanation:

⇒FOR n=1

P(1)=1³+2³+3³ is divisible by 9

⇒1+8+27 is divisible by 9

⇒36 is divisible by 9

which is true...

SUPPOSE THE STATEMENT IS TRUE FOR m=n,m∈N

P(m)=m³+(m+1)³+(m+2)³ is divisible by 9

      =m³+(m+1)³+(m+2)³=9k,k∈Z.....(i)

FOR n=m+1

P(m+1)=(m+1)³+(m+2)³+(m+3)³ is divisible by 9

now,from (i),

m³+(m+1)³+(m+2)³=9k

⇒(m+1)³+(m+2)³=9k-m³

⇒(m+1)³+(m+2)³+(m+3)³=9k-m³+(m+3)³

⇒9k-m³+m³+27+3m²(3)+3(9)m=9k+9m²+27m+27

=9(k+m²+3m+3) where k∈Z,m∈N

⇒(m+1)³+(m+2)³+(m+3)³ is divisible by 9

⇒P(m+1) is true

therefore P(m)⇒P(m+1) is true

hence,by the principle of mathematical induction the sum of cubes of three consecutive integers is divisible by 9 i.e. P(n) is true for all n∈N...

hope it helps..........pls..pls..pls..mark my ans to be the brainliest.....tq,....