Question

Find number of positive integers n such that n+2n²+3n³+....+2019n²⁰¹⁹ is divisible by natural number(n-1).

Answer 1

∵ 1 + n + n² + n³ + ...... = \(\frac{1}{n-1}\)

1 + n + n² + n³ + .... + n²⁰¹⁹ = \(\frac{1 - n^{2020}}{1-n}\)

Differentiate both sides, we get

1 + 2n + 3n² + ..... + 2019 n²⁰¹⁸ 

= \(\frac{2020(1-n)n^{2019} + (1-n^{2020})}{(1-n)^2}\)

= \(\frac{2020 n^{2019} - 2020 n^{2020} + 1-n^{2020}}{(1-n)^2}\)

multiplying both side by n, we get

n + 2n² + 3n³ + ..... + 2019 n²⁰¹⁹

= \(\frac{2020 n^{2020} - 2021 n^{2021} + n}{(1-n)^2}\)

It implies n + 2n² + .... + 2019 n²⁰¹⁹ is divisible by n-1 ∀ n∈N excepts 1