Chapter 11.2, Problem 77HP

To determine

The sum of the series $(x+y),(x+2y),(x+3y),+\mathrm{.......}$

Answer to Problem 77HP

  ${\text{S}}_{\text{n}}=nx+\frac{y}{2}(n^2+n)$

Explanation of Solution

Given:

The sequence:$(x+y),(x+2y),(x+3y),+\mathrm{.......}$

Concept Used:

The sequence:$(x+y),(x+2y),(x+3y),+\mathrm{.......}$

First term and the common difference: $a_1=x+y;d=(x+2y)-(x-y)=y$

Calculation:

  $a_1=x+y;d=y;$Find ${\text{S}}_{\text{n}}$

  $\begin{array}{l}{\text{S}}_{\text{n}}=\frac{n}{2}[2·(x+y)+(n-1)·y]\\ {\text{S}}_{\text{n}}=\frac{n}{2}[2x+2y+ny-y]\\ {\text{S}}_{\text{n}}=nx+ny+\frac{n^{2}y}{2}-\frac{ny}{2}\\ {\text{S}}_{\text{n}}=nx+\frac{n^{2}y}{2}+\frac{ny}{2}\\ {\text{S}}_{\text{n}}=nx+\frac{y}{2}(n^2+n)\end{array}$

Thus, ${\text{S}}_{\text{n}}=nx+\frac{y}{2}(n^2+n)$