Question

If P(n) is the statement" sum of first n natural numbers is divisible by n+ 1" ,prove that P(r+1) is true if P(r) is true.​

Answer 1

Step-by-step explanation:

GIVEN:P(r) is true...

⇒1+2+3+...r is divisible by r+1

let 1+2+3...r=k(r+1)

add r+1 on both sides

1+2+3...r+(r+1) = (k+1)(r+1)

                   = a multiple of r+1

Which is P(r+1)

SO if P(r) is true so is P(r+1)

                                               Q.E.D

Answer 2

Answer:

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p(n) : 3 ^ n >n^

for n= 1

p(1) : 3'1 >1^1

n=1

p(n) => 3^k >k^k

p(k+1) =3^k+1 >(k+1)^k+1

3^k.3^1>(k+1)^k ( k+1)

p(n+1) is true when p( n) is true

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