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The ruse solutions say that because they are using the results from part ii (with the t,k - t, k+1) which admittedly is worded pretty badly.

just doing the inductive step (prove true for n=k+1) LHS=(k+1c0)/x - (k+1c1)/x+1 + .... (-1)^(k+1) (k+1)c(k+1)/(x+k+1) = kc0/x - (kc1+kc0)/x+1 + (kc2+kc1)/x+2 .... (-1)^(k+1) (kck)/(x+k+1) as (kck=k+1ck+1) = k!/x(x+1)(x+2)..... - (kc0/x+1 - kc1/x+2 ......) (by assumption) now from ii we...

Hi, I would like the Emily Dickinson notes please.

Count me in please!