Question

Show that $\sum _{\mathrm{j}}^{\mathrm{n}} \left({\mathrm{a}}_{\mathrm{j}}-{\mathrm{a}}_{\mathrm{j}-1}\right)={\mathrm{a}}_{\mathrm{n}}-{\mathrm{a}}_0\text{, where}{\mathrm{a}}_0,{\mathrm{a}}_1,\dots ..{\mathrm{a}}_{\mathrm{n}}$, is a sequence of real numbers. This type of sum is called telescoping.

Short answer

Hence, proved that$\sum _{\mathrm{j}}^{\mathrm{n}} \left({\mathrm{a}}_{\mathrm{j}}-{\mathrm{a}}_{\mathrm{j}-1}\right)={\mathrm{a}}_{\mathrm{n}}-{\mathrm{a}}_0\text{, where}{\mathrm{a}}_0,{\mathrm{a}}_1,\dots ..{\mathrm{a}}_{\mathrm{n}}$ by writing out the terms in the sum.

Step-by-step solution

Step 1:

Given:$a_0,a_1,\dots .a_n$ is a sequence of real numbers

To prove that:$\sum _j^n \left(a_j-a_{j-1}\right)=a_n-a_0$

Step 2:

Let us evaluate the sum by writing out all terms in the sum:

$\begin{array}{r}\sum _{\mathrm{j}}^{\mathrm{n}} \left({\mathrm{a}}_{\mathrm{j}}-{\mathrm{a}}_{\mathrm{j}-1}\right)\\ =\left({\mathrm{a}}_1-{\mathrm{a}}_0\right)+\left({\mathrm{a}}_2-{\mathrm{a}}_1\right)+\left({\mathrm{a}}_3-{\mathrm{a}}_2\right)+\dots +\left({\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}-2}\right)+\left({\mathrm{a}}_{\mathrm{n}}-{\mathrm{a}}_{\mathrm{n}-1}\right)\end{array}$

Reorder the terms:

$=-a_0+\left(a_1-a_1\right)+\left(a_2-a_2\right)+\left(a_3-a_3\right)+\dots +\left(a_{n-2}-a_{n-2}\right)+\left(a_{n-1}-a_{n-1}\right)+a_n$

Combine like terms

$\begin{array}{r}=-{\mathrm{a}}_0+0+0+0+\dots +0+0+{\mathrm{a}}_{\mathrm{n}}\\ ={\mathrm{a}}_{\mathrm{n}}-{\mathrm{a}}_0\end{array}$