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summation of k^5 for first n natural no.s

Posted October 12th, 2008 by SD_88
in
$$(k+1)^6=k^6+6k^5+15k^4+20k^3+15k^2+6k+1$$

then you summed up for k from 1 to n
and you Reduced certain terms
and you got
(n+1)^6=1+6S_5+15S_4+20S_3+15S_2+6S_1+n

$$<br /> how did you get $$

(n+1)^6 on lhs...and summation of k^6 on rhs as 1..cud you give steps...rest all the terms are ok on rhs...shouldnt we get an S6 on rhs from the first term....
i would be thankful to you if you cud give the finl result too if possible.....or a source where it wud be present.....as i am more interested in the reswult rather than its derivation and method of getting it.....the result of summation of k^5 is of utmost importance to me now...
THANK YOU for answering me!!

‹ Arithmetic Progression, Geometric Progression & Harmonic Progression Algebra ›

Comments

Would you be thinking about

On September 28th, 2011 shawnbr28 (not verified) says:

Would you be thinking about exchanging links?

answer

On October 28th, 2008 Structure says:
$$S_5=\sum_{k=1}^{k=n}k^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$

write k+1 power 6-

On January 28th, 2009 jayasimha says:

write k+1 power 6- expansion and bring k power 6 to lhs, by giving 1,2,3... values for k we get a recurrence series and add we get sigma inlhs, and use k power 5,4,3,2,1 values in rhs an d simplify ok

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