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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.
Showthatthenumber122n102n21nisalwaysonelessthanamultipleof2020.Foreverypositiveintegern.
Commented by Prithwish sen last updated on 07/Jul/19
(101+21)^n  −(101+1)^n −21^n   ={ 101^n +n101^(n−1) 21+......+n101.(21)^(n−1) +21^n }−{101^n +n101^(n−1) +.....+n101+1)−21^n   = n101^(n−1) (21−1)+ ........+n101(21^(n−1) −1) −1  = 101×20{101^(n−2) +.......n(21^(n−2) +21^(n−2) +...+1)} −1  =2020×m −1   m=integer  which is always 1 less than multiple of 2020 . Proved
(101+21)n(101+1)n21n={101n+n101n121++n101.(21)n1+21n}{101n+n101n1+..+n101+1)21n=n101n1(211)+..+n101(21n11)1=101×20{101n2+.n(21n2+21n2++1)}1=2020×m1m=integerwhichisalways1lessthanmultipleof2020.Proved
Commented by Tawa1 last updated on 07/Jul/19
God bless you sir
Godblessyousir

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