Question

For the sequence given in Problem 103, show that

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

Short answer

It is shown that$u_{n+1}+u_n={\left(n+1\right)}^2$.

Step-by-step solution

Step 1. Write the given information.

The given information is:

$u_1=1;u_{n+1}=u_n+\left(n+1\right)$

We have to show that$u_{n+1}+u_n={\left(n+1\right)}^2$

Step 2. Find a few terms with the help of the given formula.

$$\begin{gathered} u_1=1 \\ {u}_2=u_1+\left(1+1\right) \\ =1+2 \\ =3 \\ {u}_3=u_2+\left(2+1\right) \\ =3+3 \\ =6 \\ {u}_n=u_{n-1}+\left(n-1+1\right) \\ {u}_{n-1}=u_{n-2}+\left(n-2+1\right) \end{gathered}$$

Step 3. Substitute the above-calculated values in the formula un+1=un+n+1.

^{$$\begin{gathered} u_{n+1}=u_n+\left(n+1\right) \\ =\left(u_{n-1}+\left(n-1+1\right)\right)+\left(n+1\right) \\ =\left(u_{n-2}+\left(n-2+1\right)\right)+n+\left(n+1\right) \\ =u_{n-3}+\left(n-3+1\right)+\left(n-1\right)+n+\left(n+1\right) \end{gathered}$$}

On continuing the series in the same way we will get the following series,

$$\begin{gathered} u_{n+1}=u_1+\left(2+3+....+\left(n+1\right)\right) \\ =1+2+3+....\left(n+1\right) \end{gathered}$$

The formula of first n natural numbers is: $1+2+3+....\left(n+1\right)=\frac{n\left(n+1\right)}{2}$

Therefore, $u_{n+1}=\frac{\left(n+1\right)\left(n+2\right)}{2}$

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

Step 4. Find un+1+un.

$$\begin{gathered} u_{n+1}+u_n=\frac{\left(n+1\right)\left(n+2\right)}{2}+\frac{n\left(n+1\right)}{2} \\ =\frac{\left(n+1\right)\left(n+2\right)+n\left(n+1\right)}{2} \\ =\frac{n^2+n+n^2+2n+n+2}{2} \\ =\frac{2n^2+4n+2}{2} \\ =n^2+2n+1 \\ ={\left(n+1\right)}^2 \end{gathered}$$