Show-that-the-number-122-n-102-n-21-n-is-always-one-less-than-a-multiple-of-2020-For-every-positive-integer-n-

Question Number 63674 by Tawa1 last updated on 07/Jul/19

Show that the number 122^{n} - 102^{n} - 21^{n} is always one less than a

multiple of 2020 . For every positive integer n .

Commented by Prithwish sen last updated on 07/Jul/19

{(101 + 21)}^{n} - {(101 + 1)}^{n} - 21^{n}

= {101^{n} + {n 101}^{n - 1} 21 + \dots \dots + n 101 . {(21)}^{n - 1} + 21^{n}} - {101^{n} + {n 101}^{n - 1} + \dots . . + n 101 + 1) - 21^{n}

= {n 101}^{n - 1} (21 - 1) + \dots \dots . . + n 101 (21^{n - 1} - 1) - 1

= 101 \times 20 {101^{n - 2} + \dots \dots . n (21^{n - 2} + 21^{n - 2} + \dots + 1)} - 1

= 2020 \times m - 1 \boldsymbol m = \boldsymbol integer

\boldsymbol which \boldsymbol is \boldsymbol always 1 \boldsymbol less \boldsymbol than \boldsymbol multiple \boldsymbol of 2020 . \boldsymbol Proved

Commented by Tawa1 last updated on 07/Jul/19

God bless you sir

God bless you sir

Facebook Tweet Pin