Summation problem ************

When m = 1 , ∑(k=1→n) k = n(n+1)/2 has the factor n(n+1). Assuming that for 1 ≤ m ≤ p - 1 ,
∑(k=1→n) kᵐ has the factor n(n+1) , were integer p ≥ 2.Consider
∑(k=1→n) [(k+1)ᴾ⁺¹ - kᴾ⁺¹]
= ∑(k=1→n) (k+1)ᴾ⁺¹ - ∑(k=0→n-1) (k+1)ᴾ⁺¹
= ∑(k=n) (k+1)ᴾ⁺¹ - ∑(k=0) (k+1)ᴾ⁺¹
= (n+1)ᴾ⁺¹ - 1or
∑(k=1→n) [(k+1)ᴾ⁺¹ - kᴾ⁺¹]
= ∑(k=1→n) [kᴾ⁺¹ + (p+1)kᴾ + (p+1)C2 kᴾ⁻¹ + ... + (p+1)k + 1 - kᴾ⁺¹]
= ∑(k=1→n) (p+1)kᴾ + ∑(k=1→n) (p+1)C2 kᴾ⁻¹ + ... + ∑(k=1→n) (p+1)k + n ∴
(n+1)ᴾ⁺¹ - 1
= ∑(k=1→n) (p+1)kᴾ + ∑(k=1→n) (p+1)C2 kᴾ⁻¹ + ... + ∑(k=1→n) (p+1)k + n∑(k=1→n) (p+1)kᴾ
= (n+1)ᴾ⁺¹ - (n+1) - [∑(k=1→n) (p+1)C2 kᴾ⁻¹ + ... + ∑(k=1→n) (p+1)k]Obviously , (n+1)ᴾ⁺¹ - (n+1) has the factor n+1 , further consider
(n+1)ᴾ⁺¹ - (n+1) has no constant term when we express it , therefore
(n+1)ᴾ⁺¹ - (n+1) has the factor n.Thus (n+1)ᴾ⁺¹ - (n+1) has the factor n(n+1) since n and n+1 are relative prime.By assumption ,
∑(k=1→n) (p+1)C2 kᴾ⁻¹ + ... + ∑(k=1→n) (p+1)k
has the factor n(n+1) since each terms has the factor kᵐ for 1 ≤ m ≤ p - 1.Hence ∑(k=1→n) (p+1)kᴾ has the factor n(n+1).i.e. ∑(k=1→n) kᴾ has the factor n(n+1) since p is a constant.∴ The proposition is true for m = p.By mathematical induction , the proposition is true for integer m ≥ 1.

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