Summation on HP 42S
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09-26-2018, 01:54 PM
Post: #30
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RE: Summation on HP 42S
(09-26-2018 01:23 PM)Albert Chan Wrote:(09-26-2018 12:36 PM)Gerson W. Barbosa Wrote: ... Indeed it is: '(11*n^2+48*n+49)*n/(9*((n+3)*((n+2)*(n+1))))' Well, that’s the price to pay... (09-26-2018 01:23 PM)Albert Chan Wrote: Actually, the sum is very easy, once you recognized above formula Or ∑(k=1,n,2/(k*(k+3))) = 2/3 * (11/6 - 1/(n+1) - 1/(n+2) - 1/(n+3)) Very nice! |
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