Question

The value of $\sum _{r=1}^n\left\{\left(2r-1\right)+\frac{1}{2^r}\right\}$ is _______________.

Answer 1

$$\begin{gathered} \sum _{r=1}^n\left\{\left(2r-1\right)+\frac{1}{2^r}\right\} \\ \mathrm{For}\sum _{r=1}^n\left(2r-1\right)=1+3+5+7+.......+2n+1 \end{gathered}$$
Here first term is 1

last term is 2n − 1

i.e sum is $\frac{n}{2}$ (first term + last term)
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{n}{2}\left(1+2n-1\right) \\ =\frac{n}{2}\left(2n\right) \end{gathered}$$
$$\begin{gathered} \mathrm{i}.\mathrm{e}\sum _{r=1}^n\left(2r-1\right)=n^2...\left(1\right) \\ \mathrm{also}\sum _{r=1}^n\frac{1}{2^r}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^n} \\ \mathrm{first}\mathrm{term}a=\frac{1}{2},r=\mathrm{common}\mathrm{ratio}\mathrm{is}\frac{1}{2} \\ \mathrm{Sum}{S}_{\mathit{n}}=\frac{a\left(1-r^n\right)}{1-r} \\ =\frac{{\displaystyle \frac{1}{2}\left(1-\frac{1}{2^n}\right)}}{1-{\displaystyle \frac{1}{2}}} \\ \sum _{r=1}^n\frac{1}{2^r}=S\mathrm{um}=1-\frac{1}{2^n} \\ \therefore \sum _{r=1}^n\left\{\left(2r-1\right)+\frac{1}{2^r}\right\}=n^2+1-\frac{1}{2^n} \\ \sum _{r=1}^n\left\{\left(2r-1\right)+\frac{1}{2^r}\right\}=1+n^2-\frac{1}{2^n} \end{gathered}$$