If the value of 100∑r=0(r2+4r+4)(r+1)! is (a)!−b, where 0≤b<10, then the sum of digits of a+b, is
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Solution
100∑r=0(r2+4r+4)(r+1)!=100∑r=0(r+2)(r+2)!=100∑r=0(r+3−1)(r+2)!=100∑r=0(r+3)!−100∑r=0(r+2)!=(3!+4!+⋯+103!)−(2!+3!+⋯+102!)=(103)!−2∴a+b=105
Hence sum of digits of (a+b) is 6