1.If Sr=∣2rxn(n+1)6r2−1yn2(2n+3)4r3−2nrzn3(n+1)∣S_r = \begin{vmatrix} 2r & x & n(n+1) \\ 6r^2-1 & y & n^2(2n+3) \\ 4r^3-2nr & z & n^3(n+1) \end{vmatrix}Sr=2r6r2−14r3−2nrxyzn(n+1)n2(2n+3)n3(n+1), then ∑r=1nSr\sum_{r=1}^n S_r∑r=1nSr does not depend on -a.xxxb.yyyc.nnnd.All of theseLogin to continueOnly logged in users canattempt or see the solution.
