“n(n+1)(n+2)(n+3)+1”是哪个数的平方?
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“n(n+1)(n+2)(n+3)+1”是哪个数的平方?
“n(n+1)(n+2)(n+3)+1”是哪个数的平方?
“n(n+1)(n+2)(n+3)+1”是哪个数的平方?
答:
n(n+1)(n+2)(n+3)+1
=n(n+3)*(n+1)(n+2)+1
=(n^2+3n)*(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2
所以是n^2+3n+1的平方.
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