Question 1184007
prove by mathematical induction:
1^5 + 2^5 + 3^5 + ... + n^5 = (1/12)n^2(n+1)^2(2n^2 + 2n -1)
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<pre>
Base case:  {{{1^5}}} = 1
            {{{(1/12)(1)^2(1+1)^2(2(1)^2+2(1)-1) }}} = {{{(1/12)(4)(3)}}} = 1

Base case holds.


Hypothesis:  
Assume {{{1^5 + 2^5 + 3^5}}} + ... + {{{n^5}}} = {{{ (1/12)n^2(n+1)^2(2n^2 + 2n -1) }}} for n=k.  (*)


Step case:  Let n=k+1
 [ {{{ 1^5 + 2^5 + 3^5}}} + ... + {{{k^5}}} ] + {{{(k+1)^5}}} 

... use hypothesis on terms within [  ] ...

= {{{ (1/12)k^2(k+1)^2(2k^2 + 2k -1) }}} + {{{ (k+1)^5}}}

... this reduces to (you can factor {{{(k+1)^2}}} then use a factoring website for the rest) ...  (see NOTE)

= {{{ (1/12)(k+1)^2((k+2)^2(2k^2+6k+3)) }}}

The proof is complete here, but it is not obvious, so
let u=k+1 --> k=u-1

= {{{ (1/12)(u)^2(u+1)^2(2(u-1)^2+6(u-1)+3) }}}
= {{{ (1/12)(u)^2(u+1)^2(2u^2+2u-1) }}}

Proof complete

NOTE:  I admit, there may be an easier path here.  If other tutors (or the student) wish to find it, more power to them.  I gave the general idea...

If this helps the student, thanks are appreciated.