If ∑ r = 1 10 r ! r 3 + 6 r 2 + 2 r + 5 = α 11 ! , then the value of α is equal to ___ .

If ∑r=110r!r3+6r2+2r+5=α11!, then the value of α is equal to ___ .

Solution

∑r=110r!r+1r+2r+3-9r+1+8

=∑r=110(r+3)!-(r+1)!-8r+1!-r!

=(13!+12!-2!-3!)-8(11!-1)

=(12.13+12-8)·11!-8+8

=(160)(11)!

Hence α=160

Asked in: JEE Main 2021 (18 Mar Shift 2)

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