Question

If the $1^{st}$ term and common ratio of a G.P. are $1$ and $2$ respectively then $S_1+S_3+S_5+...+S_{2n-1}=$

• A. $\left(\frac{1}{3}\right)\left(2^n-5n+4\right)$
• B. $\left(\frac{1}{3}\right)\left(2^{2n+1}-5n\right)$
• C. $\left(\frac{1}{3}\right)\left(2^{2\mathrm{n}+1}-5{\mathrm{n}}^2\right)$
• D. $\left(\frac{1}{3}\right)\left(2^{2n+1}-3n-2\right)$

Answer 1

The correct option is D

$\left(\frac{1}{3}\right)\left(2^{2n+1}-3n-2\right)$

Explanation for the Correct Option :

Step1: Apply Geometric Progression Formulas

Sum of Geometric Progression $\left(S_n\right)$:

$S_n=\frac{a(r^n-1)}{(r-1)}$, where $a$ is the first term and $r$ is common ration.

Here $a=1$and $\mathrm{r}=2$

Step2:Substitute the value of $S_n$ in the below equation

$\therefore S_1+S_3+S_5+\dots ..S_{2n-1}$

$$\begin{gathered} =1+1\frac{\left(1-2^3\right)}{1-2}+\frac{1\left(1-2^5\right)}{1-2}+.......\frac{1(1-2^{2n-1})}{1-2} \\ =1+1(2^3-1)+1(2^5-1)+\dots .+1(2^{2n-1}-1) \\ =(2+2^3+2^5+\dots ..\mathrm{n}\mathrm{terms})–(1+1+1+\dots \mathrm{n}) \\ =\frac{2(2^{2n}-1)}{(2^2-1)–n} \\ =\left(\frac{1}{3}\right)(2^{2n+1}–3n-2) \end{gathered}$$

Hence Option (3) is correct