0
$\begingroup$

Why do we get this pattern (note what has been highlighted or italicized) when you divide consecutive square numbers with consecutive odd numbers ?

4 ÷ 3 = 1 R 1

9 ÷ 5 = 1 R 4

16 ÷ 7 = 2 R 2

25 ÷ 9 = 2 R 7

36 ÷ 11 = 3 R 3

49 ÷ 13 = 3 R 10

64 ÷ 15 = 4 R 4

81 ÷ 17 = 4 R 13

100 ÷ 19 = 5 R 5

121 ÷ 21 = 5 R 16

144 ÷ 23 = 6 R 6

169 ÷ 25 = 6 R 19

196 ÷ 27 = 7 R 7

225 ÷ 29 = 7 R 22

256 ÷ 31 = 8 R 8

$\endgroup$
1
$\begingroup$

Your observation (in the bold) is that $(2n)^2$ divided by $4n-1$ has the result $n$ R $n$.

$(2n)^2=4n^2$ and if you multiply the divisor $(4n-1)$ by the quotient $n$ and add the remainder $n$ you get $((4n-1)\cdot n)+n=4n^2-n+n=4n^2$.

So that pattern is just what you expect.

Your observation (in italics) is that $(2n+1)^2$ divided by $4n+1$ has the result $n$ R $(3n+1)$.

$(2n+1)^2=4n^2+4n+1$ and if you multiply the divisor $(4n+1)$ by the quotient $n$ and add the remainder $3n+1$ you get $((4n+1)\cdot n)+3n+1=4n^2+n+3n+1=4n^2+4n+1$.

So that pattern is just what you expect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.