Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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   ....   .....    .....  ... ..... ..... ......  ...  ...   .... ....      ...  ...  1^k +2^k +3^k +...+n^k =???

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

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?

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com