Question

If 2n+1 is divisible by 3, then 10n²+n-2 is divisible by 9, where n is an element of natural numbers. prove​

Answer 1

Step-by-step explanation:

2n+1 is divisible by 3

$10 {n}^{2} + n - 2$

$= 10{n}^{2} + 5n - 4n - 2$

$= (2n + 1)(5n - 2)$

$= (2n + 1)(2n + 3n + 1 - 3)$

$= (2n + 1)(2n + 1 + 3n - 3)$

$= (2n + 1)[(2n + 1) + 3(n - 1)]$

In the above expression,

1st term, (2n+1), is divisible by 3 and

In 2nd term both (2n+1) and 3(n-1) are divisible by 3, hence 2nd term as whole is also divisible by 3

Since both the terms are divisible by 3, the whole expression will be divisible by 9.

Answer 2

Step-by-step explanation:

2n+1 is divisible by 3

10 {n}^{2} + n - 210n

2

+n−2

= 10{n}^{2} + 5n - 4n - 2=10n

2

+5n−4n−2

= (2n + 1)(5n - 2)=(2n+1)(5n−2)

= (2n + 1)(2n + 3n + 1 - 3)=(2n+1)(2n+3n+1−3)

= (2n + 1)(2n + 1 + 3n - 3)=(2n+1)(2n+1+3n−3)

= (2n + 1)[(2n + 1) + 3(n - 1)]=(2n+1)[(2n+1)+3(n−1)]

in the above expression

1st term, (2n+1), is divisible by 3 and

In 2nd term both (2n+1) and 3(n-1) are divisible by 3, hence 2nd term as whole is also divisible by 3

Since both the terms are divisible by 3, the whole expression will be divisible by 9.