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)   (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 + ... + = 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 = + ... +
...which can also be written...
= + ... + +
...apply the hypothesis to terms...
= +

...expand and simplify (lots of steps omitted here)...
=
Proof is complete, but lets show this last expression is of the form of the RHS ...
Let u = k+1 --> k=u-1:
LHS =
= (= RHS)


Answer by robertb(5830)   (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 , which is of 4th degree.
By the factor theorem, is a factor of this polynomial since .
Again by the factor theorem, is a factor of this polynomial since .
In other words, , where is a quadratic expression.
Now ===> ===> .
Also, is a 4th degree polynimial whose leading term is . Since also, this means .
===>

===> ===> ===> .
===> .

Therefore, ,
and the proof is complete.

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