Question

Solve If ${\left(1.05\right)}^{50}=11.6568$ , then $\sum _{n=1}^{49}{\left(1.05\right)}^n$ is equal to
A. 208.34
B. 213.13
C. 212.16
D. 213.16

Answer 1

Hint- In order to solve the problem start with the problem part. Use the problem statement to expand the term in the form of series. In order to find the sum first identify the type of series, use the formula for sum of the series and substitute the given value at some place in order to find the solution.

Complete step by step answer:
Given that: ${\left(1.05\right)}^{50}=11.6568$ and we have to find the value of $\sum _{n=1}^{49}{\left(1.05\right)}^n$
Let us expand the problem term that is $\sum _{n=1}^{49}{\left(1.05\right)}^n$
$$\begin{gathered} \Rightarrow \sum _{n=1}^{49}{\left(1.05\right)}^n \\ ={\left(1.05\right)}^1+{\left(1.05\right)}^2+{\left(1.05\right)}^3+.........+{\left(1.05\right)}^{49} \end{gathered}$$
As we can see every term is the product of previous term and 1.05
So the given series is a geometric progression.
Here for the given G.P.
First term $\left(a\right)=1.05$
Common ration $\left(r\right)={\displaystyle \frac{a_2}{a_1}}={\displaystyle \frac{{\left(1.05\right)}^2}{1.05}}=1.05$
Number of terms $\left(n\right)=49$
So this is a G.P. with $r>1$
As we know that the sum of first n terms of G.P. is given by:
$S_n={\displaystyle \frac{a\left(r^n-1\right)}{\left(r-1\right)}}$
As we have the values for a, r, n.
So in order to find the sum let us substitute the values in the given formula.
$$\begin{gathered} \because S_n={\displaystyle \frac{a\left(r^n-1\right)}{\left(r-1\right)}} \\ \Rightarrow S_{49}={\displaystyle \frac{1.05\left({\left(1.05\right)}^{49}-1\right)}{\left(1.05-1\right)}} \end{gathered}$$
Now let us solve the given term in order to find the sum.
$\Rightarrow S_{49}={\displaystyle \frac{{\left(1.05\right)}^{50}-1.05}{\left(1.05-1\right)}}$
Now let us substitute the value of ${\left(1.05\right)}^{50}$ from the question and solve further.
$$\begin{gathered} \Rightarrow S_{49}={\displaystyle \frac{11.6568-1.05}{\left(1.05-1\right)}} \\ \Rightarrow S_{49}={\displaystyle \frac{10.6068}{0.05}} \\ \Rightarrow S_{49}=212.157 \\ \Rightarrow S_{49}=212.16\left[\text{rounded to 2 decimal place}\right] \end{gathered}$$
Hence, the value of $\sum _{n=1}^{49}{\left(1.05\right)}^n$ is 212.16.
So, option C is the correct option.

Note- In order to solve such types of problems students must start with the problem statement as the given part in the solution is a constant and cannot be easily expanded to bring in the form of problem statement. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.