Verify that $4r^3+r=\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4$

Question

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.

Answer 1

Since $a^2-b^2=(a+b)(a-b)$, the right-hand side is$$(r^2+\tfrac14+r)^2-(r^2+\tfrac14-r)^2=(2r^2+\tfrac12)(2r)=4r^3+r.$$

Answer 2

$(x+y)^{4}-(x-y)^{3}=8x^{3}y+8xy^3$

Just substitute $x=r$ and $y=\frac{1}{2}$ to obtain

$4r^{3}+r=(r+\frac{1}{2})^{4}-(r-\frac{1}{2})^{4}$

As for the infinite sum,

$\sum_{r=1}^{n}({4r^{3}+r})=-\left( \frac{1}{2} \right)^4+\left( \frac{3}{2} \right)^4 - \left( \frac{3}{2} \right)^4+...+\left( n + \frac{1}{2} \right)^4$

$\sum_{r=1}^{n}(4r^{3}+r)=\left( n+\frac{1}{2} \right)^4-\left( \frac{1}{2} \right)^4$

Answer 3

Render

$4r^3+r=(2r^2+\frac{1}{2})(2r)$

and define

$(2r^2+\frac{1}{2})=u+v$

$2r=u-v$

Solving the latter equations for $u$ and $v$ and identifying $(u+v)(u-v)$ with $u^2-v^2$ then yields

$4r^3+r=(r^2+r+\frac14)^2-(r^2-r+\frac14)^2$

and since the arguments of the squares on the right are themselves perfect squares, this is also a difference of fourth powers:

$\color{blue}{4r^3+r=(r+\frac12)^4-(r-\frac12)^4}$

Answer 4

Can you notice that both sides of the equation are polynomials of degree less than or equal to $3$? If yes then note that if they are equal for more than $3$ values of $r$ they are equal identically.

Just put $r=0,r=1/2,r=-1/2,r=3/2$ and check equality of both sides and you are done.

The above is a general technique to prove equality of two polynomials.

Another approach is to factor the RHS as $$((r+1/2)^2+(r-1/2)^2)((r+1/2)^2-(r-1/2)^2)$$ which equals $$(2(r^2+(1/2)^2))(4r(1/2))$$ or $$(4r^2+1)r=4r^3+r$$

Answer 5

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

$$\left(n+\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4.$$

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To establish the identity, factor the differences of squares

$$\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4\\ =\left(\left(r+\frac{1}{2}\right)^2+\left(r-\frac{1}{2}\right)^2\right) \left(\left(r+\frac{1}{2}\right)+\left(r-\frac{1}{2}\right)\right)\left(\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)\right)\\ =\left(2r^2+\frac24\right)\cdot2r\cdot1.$$