Question

Prove that the product of 4 consecutive integers is 1 less than a perfect square

Answer 1

➡HERE IS YOUR ANSWER⬇

Let, the four consecutive integers are n, n+1, n+2 and n+3, where n belongs to the set of Natural numbers

Now,

n(n+1)(n+2)(n+3)

$= ( {n}^{2} + n)( {n}^{2} + 5n + 6) \\ \\ = {n}^{4} + 6 {n}^{3} + 11 {n}^{2} + 6n \\ \\ = ({n}^{4} + 6 {n}^{3} + 11 {n}^{2} + 6n + 1) - 1 \\ \\ = {(n + 1)}^{4} - 1 \\ \\ we \: \: see \: \: that \: \: n(n + 1)(n + 2)(n + 3) \: \: \\ gives \: \: a \: \: result \\ ( {(n + 1)}^{4} - 1) \: \: and \: \: {(n + 1)}^{4} \: \: is \: \: a \: \: \\ perfect \: \: square. \\ \\ so \: \: the \: \: multiplication \: \: is \: \:1 \: \: less \: \: \\ than \: \: a \: \: perfect \: \: square. \: \: (proved)$

⬆HOPE THIS HELPS YOU⬅