Solve {l}{1=11/n(n+1)=dots}{=11/(n+1)(n+2)=}

$$\left. \begin{array} { l } { 1 = 1 \frac { 1 } { n ( n + 1 ) } = \dots } \\ { = 1 \frac { 1 } { ( n + 1 ) ( n + 2 ) } = } \end{array} \right.$$

Solve for n

$$n=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$n=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Steps Using the Quadratic Formula

Divide both sides by 1.

$$1=\frac{1}{n\left(n+1\right)}$$

Variable n cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by n\left(n+1\right).

$$n\left(n+1\right)=1$$

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

$$n^{2}+n=1$$

Subtract 1 from both sides.

$$n^{2}+n-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\left(-1\right)}}{2}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$n=\frac{-1±\sqrt{1+4}}{2}$$

Add 1 to 4.

$$n=\frac{-1±\sqrt{5}}{2}$$

Now solve the equation n=\frac{-1±\sqrt{5}}{2} when ± is plus. Add -1 to \sqrt{5}.

$$n=\frac{\sqrt{5}-1}{2}$$

Now solve the equation n=\frac{-1±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from -1.

$$n=\frac{-\sqrt{5}-1}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{5}-1}{2} n=\frac{-\sqrt{5}-1}{2}$$

Steps for Completing the Square

Divide both sides by 1.

$$1=\frac{1}{n\left(n+1\right)}$$

Variable n cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by n\left(n+1\right).

$$n\left(n+1\right)=1$$

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

$$n^{2}+n=1$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 1 to \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{5}{4}$$

Factor n^{2}+n+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{\sqrt{5}}{2} n+\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$n=\frac{\sqrt{5}-1}{2} n=\frac{-\sqrt{5}-1}{2}$$