Mathematics - Determinant Question with Solution | TestHub

Using the Hockey-stick identity ${\sum }_{i=r}^n\left(\genfrac{}{}{0ex}{}{i}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r+1}\right)$, we find:$S_k=\left(\genfrac{}{}{0ex}{}{k+1}{1}\right)=k+1$ $T_k=\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)=\frac{k(k+1)}{2}$ $U_k=\left(\genfrac{}{}{0ex}{}{k+1}{3}\right)=\frac{(k-1)k(k+1)}{6}$

Substitute these into the determinant and factor out $(k+1)$, $(k+2)$, $(k+3)$ from $R_1,R_2,R_3$ respectively. Then apply row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ to simplify the determinant.$${\Delta }_k=(k+1)(k+2)(k+3)\left|\begin{array}{ccc}1& \frac{k}{2}& \frac{k(k-1)}{6}\\ 1& \frac{k+1}{2}& \frac{k(k+1)}{6}\\ 1& \frac{k+2}{2}& \frac{(k+1)(k+2)}{6}\end{array}\right|$$ $${\Delta }_k=(k+1)(k+2)(k+3)\left|\begin{array}{ccc}1& \frac{k}{2}& \frac{k(k-1)}{6}\\ 0& \frac{1}{2}& \frac{k}{3}\\ 0& 1& \frac{2k+1}{3}\end{array}\right|$$ Expanding along the first column: $${\Delta }_k=(k+1)(k+2)(k+3)\left(\frac{1}{2}\cdot \frac{2k+1}{3}-1\cdot \frac{k}{3}\right)=(k+1)(k+2)(k+3)\left(\frac{2k+1-2k}{6}\right)=\frac{(k+1)(k+2)(k+3)}{6}$$