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A sequence d(1),d(2),d(3) . . . is defin...

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`.

Text Solution

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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.
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