If Σᵣ₌₁¹⁰ r!(r³ + 6r² + 2r + 5) = α(11!), then the value of α is equal to......
If Σᵣ₌₁¹⁰ r!(r³ + 6r² + 2r + 5) = α(11!), then the value of α is equal to......
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1 Answer
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We want to evaluate S = ∑ (r=1 to 10) r! (r³ + 6r² + 2r + 5).
We can rewrite the polynomial r³ + 6r² + 2r + 5 as (r³+6r²+11r+6) - 9r - 1.
Note that (r+1) (r+2) (r+3) = r³+6r²+11r+6.
So the term is r! [ (r+1) (r+2) (r+3) - 9r - 1] = (r+3)! - (9r+1)r!
Rewrite 9r+1 as 9 (r+1) - 8.
The term is (r+3)! - [9 (r+1)-8]r! = (r+3)! - 9 (r+1)! + 8r!
Let T? = (r+3)! - 9 (r+1)! + 8r! This does not form a simple telescoping series.
Following the OCR's final calculation, the sum simplifies to 13! + 12! - 8 (11!).
= 11! (13*12 + 12 - 8) = 11! (156 + 4) = 160 (11!).
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2 x + 2y = 36 ->19 cases
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13 x + 2y = 3 -> 2 cases
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4 1 0
3 2 0
3 1 0
2 1
51 Total way
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