Can-you-Generalize-the-following-1-2-3-n-1-2-n-n-1-1-2-2-2-3-2-n-2-1-6-n-n-1-2n-1-1-3-2-3-3-3-n-3-1-2-n-n-1-2-

Question Number 2544 by Rasheed Soomro last updated on 22/Nov/15

$$\mathcal{C}{an}\:{you}\:\mathcal{G}{eneralize}\:{the}\:{following}? \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}=\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\left({n}\right)\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right. \\ $$$$\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+{n}^{\mathrm{3}} =\left[\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}} \\ $$$$….\:\:\:…..\:\:\:\:…..\:\:…\:…..\:…..\:…… \\ $$$$…\:\:…\:\:\:….\:….\:\:\:\:\:\:…\:\:… \\ $$$$\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +\mathrm{3}^{{k}} +…+{n}^{{k}} =??? \\ $$

Answered by Filup last updated on 23/Nov/15

Σ_(i=1) ^n i^k =H_n ^((−k)) Σ_(i=0) ^∞ (i+1)^k =ζ(−k) Re(k)+1<0 Σ_(i=0) ^n (i+1)^k =ζ(−k)−ζ(−k, n+2)

$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{{k}} ={H}_{{n}} ^{\left(−{k}\right)} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left({i}+\mathrm{1}\right)^{{k}} =\zeta\left(−{k}\right)\:\:\:\:\:\mathrm{Re}\left({k}\right)+\mathrm{1}<\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\left({i}+\mathrm{1}\right)^{{k}} =\zeta\left(−{k}\right)−\zeta\left(−{k},\:{n}+\mathrm{2}\right) \\ $$

Commented by Rasheed Soomro last updated on 22/Nov/15

I didn′t understand! What are m, ζ(−k) and ζ(−k, m+2) ? Formula for 1^k +2^k +...+n^k should involve only n and k. Where did m come from?

$$\mathcal{I}\:{didn}'{t}\:{understand}! \\ $$$${What}\:{are}\:{m},\:\zeta\left(−{k}\right)\:{and}\:\:\zeta\left(−{k},\:{m}+\mathrm{2}\right)\:? \\ $$$${Formula}\:{for}\:\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +…+{n}^{{k}} \:{should}\:{involve}\:{only} \\ $$$${n}\:\:{and}\:\:{k}.\:{Where}\:{did}\:{m}\:{come}\:{from}? \\ $$

Commented by prakash jain last updated on 22/Nov/15

ζ is Riemann zeta function ζ(z)=Σ_(n=1) ^∞ (1/n^z ), ℜ(z)>1

$$\zeta\:\mathrm{is}\:\mathrm{Riemann}\:\mathrm{zeta}\:\mathrm{function} \\ $$$$\zeta\left({z}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{z}} },\:\Re\left({z}\right)>\mathrm{1} \\ $$

Commented by Filup last updated on 23/Nov/15

Sorry, I did a direct copy from Wolfram without adjusting for your variables. I′ve fixed it to suite your question!

$${Sorry},\:{I}\:\mathrm{did}\:\mathrm{a}\:\mathrm{direct}\:\mathrm{copy}\:\mathrm{from} \\ $$$$\mathrm{W}{olfram}\:\mathrm{without}\:\mathrm{adjusting}\:\mathrm{for}\: \\ $$$$\mathrm{your}\:\mathrm{variables}.\:\mathrm{I}'{ve}\:\mathrm{fixed}\:\mathrm{it}\:\mathrm{to}\:\mathrm{suite} \\ $$$$\mathrm{your}\:\mathrm{question}! \\ $$

Commented by Rasheed Soomro last updated on 23/Nov/15

Could you answer in simplified form. I mean without using riemann zeta notation?

$$\boldsymbol{\mathrm{C}}\mathrm{ould}\:\mathrm{you}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{simplified}\:\mathrm{form}.\:\mathrm{I}\:\mathrm{me}{an}\: \\ $$$$\mathrm{without}\:\mathrm{using}\:\:\mathrm{riemann}\:\mathrm{zeta}\:\mathrm{notation}? \\ $$

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