Product of two consecutive positive integers is divisible by 2
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USING PMI
so we have to prove n(n+1) is divisible by 2 where n ∈ I
for n = 1
1 x 2 = 2 which is divisible by 2 so we see of n = 1 the statement is true
so let n(n+1) = 2α ⇒ n² + n = 2α
so the statement is true if n = n+1 is also divisible by 2
putting n = n+1
we get (n+1)(n+2) = n² + n + 2n + 2 = 2α + 2n + 2 (using n² + n = 2α )
= 2(α + n + 1)
so we see for n = n+1 is also divisible by 2 so the statement product of two consecutive integers is always divisible by 2 is true
so we have to prove n(n+1) is divisible by 2 where n ∈ I
for n = 1
1 x 2 = 2 which is divisible by 2 so we see of n = 1 the statement is true
so let n(n+1) = 2α ⇒ n² + n = 2α
so the statement is true if n = n+1 is also divisible by 2
putting n = n+1
we get (n+1)(n+2) = n² + n + 2n + 2 = 2α + 2n + 2 (using n² + n = 2α )
= 2(α + n + 1)
so we see for n = n+1 is also divisible by 2 so the statement product of two consecutive integers is always divisible by 2 is true
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