A sequence `d_(1),d_(2),d_(3) . . .` is defined by letting `d_(1)=2` and `d_(k)=(d_(k-1))/(k),` for all natural numbers, `k≥2`. Show that `d_(n)=(2)/(n!)`, for all `n in N`.

A sequence `d_(1),d_(2),d_(3) . . .` is defined by letting `d_(1)=2` a

1.	A sequence `d_(1),d_(2),d_(3) . . .` is defined by letting `d_(1)=2` and `d_(k)=(d_(k-1))/(k),` for all natural numbers, `k≥2`. Show that `d_(n)=(2)/(n!)`, for all `n in N`.
Answer» Let P(n) : `d_(n)=(2)/(n!)AAninN`, to prove P(2) is true. Step I `P(2):d_(2)=(2)/(2!)=(2)/(2xx1)=1` As, given `d_(1)=2` `rArrd_(k)=(d_(k-1))/(k)` `rArrd_(2)=(d_(1))/(2)=(2)/(2)=1` Hence, P(2) is true Step II Now, assume that P(k) is true. `P(k):d_(k)=(2)/(k!)` Step III Now, to prove that P(k+1) is true, we have to show that `P(k+1):d_(k+1)=(2)/(k+1)!` `d_(k+1)=(d_(k+1-1))/(k)=(d_(k))/(k)` `=(2)/(k!k)=(2)/(k+1)!` So, P(k+1) is true. Hence, P(n) is true.