<s>[INST] Verify that $4r^3+r=\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4$ Must I verify that Left hand side(LHS)= Right hand side(RHS)or can I prove that RHS= LHS?
I don’t know how to prove from LHS=RHS. How to separate the $4r^3+r$ into two terms, i.e. $\displaystyle\Bigl(r+\frac{1}{2}\Bigr)^4-\Bigl(r-\frac{1}{2}\Bigr)^4$
After verifying, I will need to find $$\sum_{r=1}^n (4r^3+r)$$
This is a summation of finite series question.
 [/INST] Using the first identity,
$$\sum_{r=1}^n(4r^3+r)=\sum_{r=1}^n\left(r+\frac{1}{2}\right)^4-\sum_{r=1}^n\left(r-\frac{1}{2}\right)^4
\\=\sum_{r=1}^n\left(r+\frac{1}{2}\right)^4-\sum_{r=0}^{n-1}\left(r+\frac{1}{2}\right)^4$$
and by telescoping, this is
$$\left(n+\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4.$$

To establish the identity, factor the differences of squares
$$\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4\\
=\left(\left(r+\frac{1}{2}\right)^2+\left(r-\frac{1}{2}\right)^2\right)
\left(\left(r+\frac{1}{2}\right)+\left(r-\frac{1}{2}\right)\right)\left(\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)\right)\\
=\left(2r^2+\frac24\right)\cdot2r\cdot1.$$
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