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Question Number 196085 by sonukgindia last updated on 17/Aug/23

Answered by MM42 last updated on 17/Aug/23

(1^(17) +2023^(17) )+(2^(17) +2022^(17) )+...+1012^(17) =  (1+2023)k_1 +(2+2022)k_2 +...+2^(17) ×506^(17)   =8k≡^8 0

$$\left(\mathrm{1}^{\mathrm{17}} +\mathrm{2023}^{\mathrm{17}} \right)+\left(\mathrm{2}^{\mathrm{17}} +\mathrm{2022}^{\mathrm{17}} \right)+...+\mathrm{1012}^{\mathrm{17}} = \\ $$$$\left(\mathrm{1}+\mathrm{2023}\right){k}_{\mathrm{1}} +\left(\mathrm{2}+\mathrm{2022}\right){k}_{\mathrm{2}} +...+\mathrm{2}^{\mathrm{17}} ×\mathrm{506}^{\mathrm{17}} \\ $$$$=\mathrm{8}{k}\overset{\mathrm{8}} {\equiv}\mathrm{0} \\ $$

Answered by qaz last updated on 18/Aug/23

Σ_(k=1) ^(2023) k^(17) =(1/2)Σ_(k=1) ^(2023) (k^(17) +(2024−k)^(17) )≡(1/2)Σ_(k=1) ^(2023) (k^(17) +(−k)^(17) )≡0(mod 8)

$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}{k}^{\mathrm{17}} =\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left({k}^{\mathrm{17}} +\left(\mathrm{2024}−{k}\right)^{\mathrm{17}} \right)\equiv\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\left({k}^{\mathrm{17}} +\left(−{k}\right)^{\mathrm{17}} \right)\equiv\mathrm{0}\left({mod}\:\mathrm{8}\right) \\ $$

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