Most Important Selected Qs for JEE AdvancedMathematicsSequences and Series

If _ r=1 ¹⁰⁰³ (r^2+1 ) r!=a!-b c! where a, b, c N , then find the least value of (a+b+c)

Correct answer

2011

Step-by-step solution

We have _ r=1 ¹⁰⁰³ (r^2+1 ) r!= _ r=1 ¹⁰⁰³ (r^2+r )-(r-1) r! aligned & = _ r=1 ¹⁰⁰³ r(r+1) r!-(r-1) r! = _ r=1 ¹⁰⁰³ r (r+1)!-(r-1) r! & =(1 2!-0)+(2 3!-1 2!)+(3 4!-2 3!)+ +(1003 1004!-1002 1003!) & =1003 1004!=(1005-2) 1004!=1005 1004!-2 1004! & =1005!-2 1004!=a-b c! & Hence (a+b+c)_ least =1005+2+1004=2011 aligned

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