prove that n2+n is divisible by 2 for evry positive integer n.
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696
Here is the answer to your question.
Any positive integer is of the form 2q or 2q + 1, where q is some integer
When n = 2q,
n²+n=(2q)²+2q
=4q²+2q
= 2q(2q+1)
which is divisible by 2
when n=2q+1
n²-n=(2q+1)²+(2q+1)
= 4q²+4q+1+2q+1
=4q²+6q +2
= 2(2q²+3q+1)
which is divisible by 2
hence n²+n is divisible by 2 for every positive integer n
Any positive integer is of the form 2q or 2q + 1, where q is some integer
When n = 2q,
n²+n=(2q)²+2q
=4q²+2q
= 2q(2q+1)
which is divisible by 2
when n=2q+1
n²-n=(2q+1)²+(2q+1)
= 4q²+4q+1+2q+1
=4q²+6q +2
= 2(2q²+3q+1)
which is divisible by 2
hence n²+n is divisible by 2 for every positive integer n
anustarnoor:
i have done
thanks bhai
can u solve one more ans
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tell
show that the square of an odd positive integers is of the form 8m+1, for some whole number m.
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