# S=\frac{d_1}{1+k}+\frac{d_1(1+g)}{(1+k)^2}+\frac{d_1(1+g)}{(1+k)^3}+...and k > g a....

## Question:

S={eq}\frac{d_1}{1+k}+\frac{d_1(1+g)}{(1+k)^2}+\frac{d_1(1+g)}{(1+k)^3}+ {/eq}...and k > g

a. {eq}\frac{d_1}{k-g} {/eq}

b. {eq}\frac{d_1}{1+k} {/eq}

c. {eq}\frac{d_1(1-(\frac{1+g}{1+k})^n}{k-g} {/eq}

d.{eq}\frac{d_1(1-(\frac{1+g}{1+k})^n+1}{k-g} {/eq}

e. None of the above

## Geometric Series:

A geometric series is a series where the ratio between consecutive terms is constant. Whether a geometric series converges or not depends on the value of the ratio.

## Answer and Explanation:

The answer is a).

The series in this question is a geometric series. In general, for a geometric series with a first term {eq}a {/eq}, a constant ratio {eq}g < 1{/eq}, the sum of the series is given by:

- {eq}\dfrac{a }{1 - g} {/eq}

In this question, the first term is {eq}\dfrac{d_1}{1 + k}{/eq}, the constant ratio is {eq}\dfrac{1 + g}{1 + k}{/eq}, thus the sum of the series is:

- {eq}\dfrac{\dfrac{d_1}{1 + k}}{1 - \dfrac{1 + g}{1 + k}} = \dfrac{\dfrac{d_1}{1 + k}}{\dfrac{k - g}{1 + k}} = \dfrac{d_1}{k - g} {/eq}

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