4 years ago

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

Mathematics

2 answers:

RoseWind [281]4 years ago

8 0

Answer:

2016

Step-by-step explanation:

alisha [4.7K]4 years ago

5 0

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

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