Solve (n(n+1)/2)^2=4k

$$( \frac { n ( n + 1 ) } { 2 } ) ^ { 2 } = 4 k$$

Solve for k

$$k=\frac{\left(n\left(n+1\right)\right)^{2}}{16}$$

Steps for Solving Linear Equation

Use the distributive property to multiply n by n+1.

$$\left(\frac{n^{2}+n}{2}\right)^{2}=4k$$

To raise \frac{n^{2}+n}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(n^{2}+n\right)^{2}}{2^{2}}=4k$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n^{2}+n\right)^{2}.

$$\frac{\left(n^{2}\right)^{2}+2n^{2}n+n^{2}}{2^{2}}=4k$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{n^{4}+2n^{2}n+n^{2}}{2^{2}}=4k$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{n^{4}+2n^{3}+n^{2}}{2^{2}}=4k$$

Calculate 2 to the power of 2 and get 4.

$$\frac{n^{4}+2n^{3}+n^{2}}{4}=4k$$

Divide each term of n^{4}+2n^{3}+n^{2} by 4 to get \frac{1}{4}n^{4}+\frac{1}{2}n^{3}+\frac{1}{4}n^{2}.

$$\frac{1}{4}n^{4}+\frac{1}{2}n^{3}+\frac{1}{4}n^{2}=4k$$

Swap sides so that all variable terms are on the left hand side.

$$4k=\frac{1}{4}n^{4}+\frac{1}{2}n^{3}+\frac{1}{4}n^{2}$$

The equation is in standard form.

$$4k=\frac{n^{4}}{4}+\frac{n^{3}}{2}+\frac{n^{2}}{4}$$

Divide both sides by 4.

$$\frac{4k}{4}=\frac{n^{2}\left(n+1\right)^{2}}{4\times 4}$$

Dividing by 4 undoes the multiplication by 4.

$$k=\frac{n^{2}\left(n+1\right)^{2}}{4\times 4}$$

Divide \frac{n^{2}\left(1+n\right)^{2}}{4} by 4.

$$k=\frac{n^{2}\left(n+1\right)^{2}}{16}$$

Solve for n (complex solution)

$$n=\frac{-\sqrt{16\sqrt{k}+1}-1}{2}$$

$$n=\frac{\sqrt{16\sqrt{k}+1}-1}{2}$$

$$n=\frac{\sqrt{-16\sqrt{k}+1}-1}{2}$$

$$n=\frac{-\sqrt{-16\sqrt{k}+1}-1}{2}$$

Solve for n

$$\left\{\begin{matrix}n=\frac{-\sqrt{16\sqrt{k}+1}-1}{2}\text{; }n=\frac{\sqrt{16\sqrt{k}+1}-1}{2}\text{, }&k\geq 0\\n=\frac{-\sqrt{-16\sqrt{k}+1}-1}{2}\text{; }n=\frac{\sqrt{-16\sqrt{k}+1}-1}{2}\text{, }&k\geq 0\text{ and }k\leq \frac{1}{256}\end{matrix}\right.$$