A) depends only on a
B) depends only on n
C) depends both on a and n
D) is independent of both a and n
Correct Answer: D
Solution :
\[\sum\limits_{r=1}^{n-1}{r=1+2+3+...+(n-1)=\frac{n(n-1)}{2}}\] \[\sum\limits_{r=1}^{n-1}{(2r=1)=1+3+5+...+[2(n-1)-2]}\] \[={{(n-1)}^{2}}\] \[\sum\limits_{r=1}^{n-1}{(3r-1)=1+4+7+...+(3n-3-2)}\] \[=\frac{(n-1)(3n-4)}{2}\] \[\therefore \]\[\sum\limits_{r=1}^{n-1}{{{\Delta }_{r}}}=\left| \begin{matrix} \Sigma r & \Sigma (2r-1) & \Sigma (3r-2) \\ \frac{n}{2} & n-1 & a \\ \frac{n(n-1)}{2} & {{(n-1)}^{2}} & \frac{(n-1)(3n-4)}{2} \\ \end{matrix} \right|\] \[\sum\limits_{r=1}^{n-1}{{{\Delta }_{r}}}\]consists of (n ? 1) determinants in L.H.S. and in R.H.S every constituent of first row consists of (n ? 1) elements and hence it can be splitted into sum of (n ? 1) determinants. \[\therefore \]\[\sum\limits_{r=1}^{n-1}{{{\Delta }_{r}}}=\left| \begin{matrix} \frac{n(n-1)}{2} & {{(n-1)}^{2}} & \frac{(n-1)(3n-4)}{2} \\ \frac{n}{2} & n-1 & a \\ \frac{n(n-1)}{2} & {{(n-1)}^{2}} & \frac{(n-1)(3n-4)}{2} \\ \end{matrix} \right|\] \[=0\] (\[\because \]\[{{R}_{1}}\]and \[{{R}_{3}}\] are identical) Hence, value of\[\sum\limits_{r=1}^{n-1}{{{\Delta }_{r}}}\]is independent of both 'a' and 'n'.
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