Đáp án+Giải thích các bước giải:
`A=n(n+1)(n+2)(n+3)+1`
`=(n^2+3n)(n^2+n+2n+2)+1`
`=(n^2+3n)(n^2+3n+2)+1`
Đặt `y=n^3+3n` nên:
`A=y(y+2)+1`
`=y^2+2y+1`
`=(y+1)^2`
Hay `A=(n^2+3n+1)^2`
Vì `n∈N=>(n^2+3n+1)^2∈N`
`=>A ` là số chính phương
~ Bạn tham khảo ~
Ta có :
`A = n(n+1)(n+2)(n+3)+1`
`=> A = (n^2+3n)(n^2+3n+2)+1`
`=> A = (n^2 + 3n + 1 - 1)(n^2 + 3n + 1 + 1) + 1`
`=> A = [(n^2 + 3n + 1) - 1][(n^2 + 3n + 1) + 1] + 1`
`=> A = (n^2+3n+1)^2 - 1+1`
`=> A = (n^2+3n+1)^2`
Vậy `A` là số chính phương
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