Question

$\underset{x\to 3}{\lim }\frac{{\displaystyle \sum _{r=1}^n}{x}^r-{\displaystyle \sum _{r=1}^n}{3}^r}{x-3}$ is real to

(a) $\frac{\left(2n-1\right)\times 3^n}{4}$

(b) $\frac{\left(2n-1\right)\times 3^n+1}{4}$

(c) $\left(2n-1\right)3^n+1$

(d) $\frac{\left(2n-1\right)\times 3^n-1}{4}$

Answer 1

(b) $\frac{\left(2n-1\right)\times 3^n+1}{4}$

$$\begin{gathered} \underset{x\to 3}{\lim }\frac{{\displaystyle \sum _{r=1}^n}{x}^r-{\displaystyle \sum _{r=1}^n}{3}^r}{x-3} \\ =\underset{x\to 3}{\lim }\frac{x^1+x^2+x^3+......+x^n-\left(3^1+3^2+3^3.....3^n\right)}{x-3} \\ =\underset{x\to 3}{\lim }\frac{x-3}{x-3}+\frac{x^2-3^2}{x-3}+\frac{x^3-3^3}{x-3}+....\frac{x^n-3^n}{x-3} \\ =1+2\times 3+3\times 3^2+.....+n{3}^{n-1}\left[\because \frac{x^n-a^n}{x-a}=n{a}^{n-1}\right] \end{gathered}$$

This series is an AGP of the form given below:

S = 1 + 2r + 3r²........nr^n–1
$$\begin{gathered} \mathrm{S}=\frac{1-r^n}{{\left(1-r\right)}^2}-\frac{n{r}^n}{1-r} \\ r=3,a=1d=1 \\ =\frac{1-3^n+2n{3}^n}{4} \\ =\frac{3^n\left(2n-1\right)+1}{4} \end{gathered}$$