Question

Using the method of telescopic sums, solve 

the recurrence relation : (n+1)x_n-nx_n-1-n/2,x₀=10,n≥1

Answer 1

$(n+1)x_n-nx_{n-1}=n/2$

Let $nx_{n-1}=y_n$. Then $y_{n+1}-y_n=n/2$ and $y_1=x_0=10$.

$y_n=(y_n-y_{n-1})+(y_{n-1}-y_{n-2})+\dots+(y_2-y_1)+y_1=$

$=\frac{n-1}{2}+\frac{n-2}{2}++\dots+\frac{1}{2}+10=10+\frac{n(n-1)}{4}$

$x_{n-1}=y_n/n=\frac{10}{n}+\frac{n-1}{4}$

$x_n=\frac{10}{n+1}+\frac{n}{4}$

Check the solution:

$x_0=\frac{10}{0+1}+\frac{0}{4}=10$

$(n+1)x_n-nx_{n-1}=(n+1)(\frac{10}{n+1}+\frac{n}{4})-n(\frac{10}{n}+\frac{n-1}{4})=$

$10+\frac{n(n+1)}{4}-(10+\frac{n(n-1)}{4})=n/2$