1.

Let Δr=∣∣∣∣∣r−1n12(r−1)22n28n−4(r−1)33n36n2−6n∣∣∣∣∣. Then the value of n∑r=1Δr is:

Answer»

Let Δr=

r1n12(r1)22n28n4(r1)33n36n26n

. Then the value of nr=1Δr is:



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