Posted on: Wed, 2007-11-14 18:27
tricky problem
Find the sum 1/2!17! + 1/3!16! + 1/4!15! + 1/5!14! ...........+ 1/9!10!
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CAT4MBA |
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Posted on: Wed, 2007-11-14 18:27
tricky problem
Find the sum 1/2!17! + 1/3!16! + 1/4!15! + 1/5!14! ...........+ 1/9!10!
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What are the options. . .
is it
Find the sum 1/(2!17!) + 1/(3!16!) + 1/(4!15!) + 1/5!14! ...........+ 1/(9!10!)
= 1/19! [ 19C2 + 19C3 + 19C4 + . . . . . +19C9 ]
19C2 + 19C3 + 19C4 + . . . . . +19C9 = 19C17 + 19C16 + 19C15 + . . . . . +19C10
= 1/2*19![19C2 + 19C3 + 19C4 + . . . . . +19C9 + 19C17 + 19C16 + 19C15 + . . . . . +19C10 ]
=(217-1)/2*19!
n/a
= 1/2*19![19C2 + 19C3 + 19C4 + . . . . . +19C9 + 19C17 + 19C16 + 19C15 + . . . . . +19C10 ]
=1/2*19![2(19C0+19C1)+19C2 + 19C3 + 19C4 + . . . . . +19C9 +19C17 + 19C16 + 19C15 + . . . . . +19C10 - 40]
=1/2*19![219 - 40]
n/a
Ans. (218 - 20 )/19!
Can you please explain how you found the answer
the series can be rewritten as 1/19!(19C0 + 19C1 + 19C2 + 19C3...+19C9 - 19C0-19C1) => 1/19!((2^19/2))-1-19) (bcos 19C0 + 19C1 + 19C2 +...19C19 = 2^19) AND 19C0+..19C9 is the halfway after that it starts repeating) therefore we get 1/19!((2^19/2))-1-19) => (2^18-20)/19!
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