Question

If $A_r=\left|\begin{array}{ccc}1& r& 2^r\\ 2& n& n^2\\ n& \frac{n\left(n+1\right)}{2}& 2^{n+1}\end{array}\right|$, then the value of $\sum _{r=1}^n{A}_r$ is
(a) n
(b) 2n
(c) − 2n
(d) n²

Answer 1

$$\begin{gathered} A_r=\left|1r{2}^r \\ 2n{n}^2 \\ n\frac{n\left(n+1\right)}{2}{2}^{n+1}\right| \\ \Rightarrow \sum _{r=1}^n{A}_r=\left|\sum _{r=1}^{n}1\sum _{r=1}^{n}r\sum _{r=1}^n{2}^r \\ \sum _{r=1}^{n}2n{n}^2 \\ n\frac{n\left(n+1\right)}{2}{2}^{n+1}\right| \\ As\sum _{r=1}^{n}1=1+1+1...+1(ntimes)=n \\ \sum _{r=1}^{n}r=1+2+3+...+n=\frac{n\left(n+1\right)}{2} \\ \mathrm{Let}S=\sum _{r=1}^n{2}^r=2+2^2+2^3=...+2^n \\ 2S=2^2+2^3=...+2^n+2^{n+1} \\ \Rightarrow 2S-S=S=\sum _{r=1}^n{2}^r=2^{n+1}-2 \\ \Rightarrow \sum _{r=1}^n{A}_r=\left|n\frac{n\left(n+1\right)}{2}{2}^{n+1}-2 \\ 2nn{n}^2 \\ n\frac{n\left(n+1\right)}{2}{2}^{n+1}\right| \\ \left[\mathrm{Applying}{R}_1\to R_1-R_2\right] \\ \sum _{r=1}^n{A}_r=\left|n-n\frac{n\left(n+1\right)}{2}-\frac{n\left(n+1\right)}{2}{2}^{n+1}-2-2^{n+1} \\ 2nn{n}^2 \\ n\frac{n\left(n+1\right)}{2}{2}^{n+1}\right| \\ =\left|00-2 \\ 2nn{n}^2 \\ n\frac{n\left(n+1\right)}{2}{2}^{n+1}\right| \\ =-2\times \left|2nn \\ n\frac{n\left(n+1\right)}{2}\right| \\ =-2\left[n^3+n^2-n^2\right] \\ =-2n^3 \end{gathered}$$