1.

Let Δr=∣∣∣∣∣r−1n6(r−1)22n24n−2(r−1)33n33n2−3n∣∣∣∣∣. Then the value of n∑r=1Δr is:

Answer»

Let Δr=

r1n6(r1)22n24n2(r1)33n33n23n

. Then the value of nr=1Δr is:



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