SOLUTION: prove using mathematical induction: 1. 1^4 + 2^4 + 3^4 + ... + n^4 = (1/30)n(n+1)(2n+1)(3n^2 + 3n - 1)

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Question 1184006: prove using mathematical induction:
1. 1^4 + 2^4 + 3^4 + ... + n^4 = (1/30)n(n+1)(2n+1)(3n^2 + 3n - 1)





Found 2 solutions by math_helper, robertb:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
prove using mathematical induction:
1. 1^4 + 2^4 + 3^4 + ... + n^4 = (1/30)n(n+1)(2n+1)(3n^2 + 3n - 1)
-----------------------



Base case, n=1:   LHS = 1^4 = 1

                  RHS = (1/30)(1*2*3*5) = (1/30)(30) = 1   (base case holds) 





Hypothesis:

Assume +1%5E4+%2B+2%5E4+%2B+3%5E4+ + ... + n%5E4 = +%281%2F30%29n%28n%2B1%29%282n%2B1%29%283n%5E2+%2B+3n+-+1%29+   for  n=k





Step case:  Let n=k+1

Thus far, it has been setup.  The task now is to show LHS = RHS for n=k+1, and the proof will be complete.



LHS = +1%5E4+%2B+2%5E4+%2B+3%5E4+ + ... + +%28k%2B1%29%5E4+ 



...which can also be written...



= +green%28+1%5E4+%2B+2%5E4+%2B+3%5E4%29+ + ... + +green%28k%5E4%29+ + +%28k%2B1%29%5E4+ 



...apply the hypothesis to green%28green%29 terms...



= +%281%2F30%29k%28k%2B1%29%282k%2B1%29%283k%5E2+%2B+3k+-+1%29+ + +%28k%2B1%29%5E4+ 





...expand and simplify (lots of steps omitted here)...

= +%281%2F30%29%28k%2B1%29%28k%2B2%29%282k%2B3%29%283k%5E2%2B9k%2B5%29+



Proof is complete, but lets show this last expression is of the form of the RHS +%281%2F30%29n%28n%2B1%29%282n%2B1%29%283n%5E2+%2B+3n+-+1%29+ ...



Let u = k+1  -->  k=u-1:



LHS =  +%281%2F30%29%28u%29%28u%2B1%29%282%28u-1%29%2B3%29%283%28u-1%29%5E2%2B9%28u-1%29%2B5%29+

    =  +%281%2F30%29u%28u%2B1%29%282u%2B1%29%283u%5E2%2B3u-1%29+  (= RHS)

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
I will go directly to step 3 of the Inductive process, and will assume the Inductive hypothesis to be true, i.e.,
for some n = k.
Prove: is true for n = k + 1, that is,
.

From the inductive hypothesis,


Focus on the the polynomial red%28p%28k%29%29+=+k%282k%2B1%29%283k%5E2+%2B+3k+-+1%29+%2B+30%28k%2B1%29%5E3 , which is of 4th degree.
By the factor theorem, k%2B2 is a factor of this polynomial since .
Again by the factor theorem, 2k%2B3 is a factor of this polynomial since .
In other words, red%28p%28k%29%29+=+red%28+%28k%2B2%29%282k%2B3%29p%5B2%5D%28k%29%29 , where p%5B2%5D%28k%29+=+ak%5E2%2Bbk+%2Bc is a quadratic expression.
Now ===> 30+=+6c ===> c=5.
Also, p%28k%29 is a 4th degree polynimial whose leading term is 6k%5E4. Since p%28x%29+=+%28k%2B2%29%282k%2B3%29%28ak%5E2%2Bbk+%2Bc%29 also, this means a+=+3.
===>

===> ===> 3%2A5+%2B30%2A8+=+3%2A5%2A%28b%2B8%29 ===> b=9.
===> .

Therefore, ,
and the proof is complete.