Prove that 2017(2018^9+2018^8+…+2018^2+2019+1)=2018^10+2017

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asked Sep 11, 2020 136k views

4 votes

Prove that 2017(2018^9+2018^8+…+2018^2+2019+1)=2018^10+2017

Mathematics
middle-school

Jiten asked

by Jiten

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3 votes

Answer:

2016

Explanation:

Kajarigd answered Sep 11, 2020

by Kajarigd

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2 votes

let a=2018, we can rewrite the question into

$(a-1)(a^9+a^8+...+a^2+a+2)\\=(a-1)(a^9+a^8+...+a^2+a+1)+(a-1)\\=a^(10)-1+a-1$

substitute back a = 2018, obtained

${2018}^(10) + 2016$
well, there is something wrong here

Gina answered Sep 18, 2020

by Gina

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