Question-if-k-is-odd-amp-A-1-k-2-k-n-k-amp-B-1-2-n-prove-that-B-A-

Question Number 192425 by mehdee42 last updated on 17/May/23

Q u e s t i o n

i f “ k^{″} i s o d d & A = 1^{k} + 2^{k} + \dots + n^{k} & B = 1 + 2 + \dots + n

p r o v e t h a t : B ∣ A

Answered by MM42 last updated on 24/May/23

W e k h n o w; 1 + 2 + \dots + n = \frac{n (n + 1)}{2} & (n, \frac{n + 1}{2}) = 1 o r (\frac{n}{2}, n + 1) = 1

T h u s, i t s o f f i c e s t o s h o w t h a t

n ∣ A & \frac{n + 1}{2} ∣ A o r \frac{n}{2} ∣ A & n + 1 ∣ A

p r o o f ∴

i f “ n^{″} i s e v e n \Rightarrow

\frac{n}{2} ∣ 1^{k} + {(n - 1)}^{k}, 2^{k} + {(n - 2)}^{k}, \dots (i)

n + 1 ∣ 1^{k} + n^{k}, 2^{k} + {(n - 1)}^{k}, \dots (i i)

(i), (i i) \Rightarrow B ∣ A

i f “ n^{″} i s o d d \Rightarrow

n ∣ 1^{k} + {(n - 1)}^{k}, 2^{k} + {(n - 2)}^{k}, \dots (i i i)

\frac{n + 1}{2} ∣ 1^{k} + n^{k}, 2^{k} + {(n - 1)}^{k}, \dots (i v)

(i i i), (i v) \Rightarrow B ∣ A

\Rightarrow \forall n \in N \Rightarrow 1 + 2 + 3 + \dots + n ∣ 1^{k} + 2^{k} + \dots + n^{k}; i f “ k^{″} i s “ {o d d}^{″}

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