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**1. Homework Statement**

Prove:

[tex]1^{4}+2^{4}+3^{4}+...+n^{4}=\frac{(n)(n+1)(2n+1)(3n^{2}+3n-1)}{30}[/tex]

**2. Homework Equations**

Umm, Im not using any.

**3. The Attempt at a Solution**

So my first step:

1) Check for n=1

[tex]1^{4}=\frac{1(1+1)(2(1)+1)(3(1)^{2}+3(1)-1)}{30}=\frac{(2)(3)(5)}{30}=1[/tex]

2)Now if n=k

[tex]P_{k}=1^{4}+2^{4}+3^{4}+...+k^{4}=\frac{(k)(k+1)(2k+1)(3k^{2}+3k-1)}{30}[/tex] is assumed true.

So, when n=k+1

So I would be expecting to get: [tex]P_{k+1}:1^{4}+2^{4}+3^{4}+...+k^{4}+(k+1)^{4}=\frac{(k+1)(k+2)(2k+3)(3(k+1)^{2}+3k+3-1)}{30}[/tex]

Since we assumed Pk to be true, and [tex]P_{k}+(k+1)^{4}=P_{k+1}[/tex]

Lets prove it. The Left Hand Side (LHS) is the same, but the RHS is giving me trouble.

I am supposed to obtain

[tex]\frac{(k+1)(k+2)(2k+3)(3k^{2}+9k+6)}{30}[/tex]

By doing something to: [tex]\frac{(k)(k+1)(2k+1)(3k^{2}+3k-1)}{30}+(k+1)^{4}[/tex]

So my first try, i'll take out (k+1) as a common factor.

So: [tex](k+1)(\frac{(k)(2k+1)(3k^{2}+3k-1)}{30}+(k+1)^{3})[/tex]

Any hints on what to do now :S?

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