Question

Find the number of positive integers ‘n’ such that
n + 2n^(2) + 3n^(3) + ……….. + 2020n^(2020) is divisible by (n – 1) (n > 1).

Answer 1

Step-by-step explanation:

1 + n + n2 + n3 + ...... = 1 n − 1 1n−1 1 + n + n2 + n3 + .... + n2019 = 1 − n 2020 1 − n 1−n20201−n Differentiate both sides, we get 1 + 2n + 3n2 + ..... + 2019 n2018 = 2020 ( 1 − n ) n 2019 + ( 1 − n 2020 ) ( 1 − n ) 2 2020(1−n)n2019+(1−n2020)(1−n)2 = 2020 n 2019 − 2020 n 2020 + 1 − n 2020 ( 1 − n ) 2 2020n2019−2020n2020+1−n2020(1−n)2 multiplying both side by n, we get n + 2n2 + 3n3 + ..... + 2019 n2019 = 2020 n 2020 − 2021 n 2021 + n ( 1 − n ) 2 2020n2020−2021n2021+n(1−n)2 It implies n + 2n2 + .... + 2019 n2019 is divisible by n-1 ∀ n∈N