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Grade: 11
        If 1.(0!) + 3(1!) + 7(2!) + 13(3!) + 21(4!)+... upto n terms = (4000)4000!, then n is equal to ????
8 months ago

Answers : (1)

Arun
13047 Points
							
Dear student
 
Remember that
Sigma[k! [(k+1)(k+2) - 2(k+1) + 1 ]
 = (k+1) [(k+1)! ]
 
Hence in the above question 
k = 3999
 
Regards
Arun (askIITisns forum expert)
8 months ago
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