Answer Part (b)?

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Answer 1 · 2018-03-22T07:51:15

See below.

If

#a_n = 2n^2-3n+5# then

#a_(n+1) = 2(n+1)^2-3(n+1)+5 = 2n^2+4n+2-3n-3+5 = a_n + 4n-1 = 4(n+1)-5#

but we know that

#a_n = a_(n-1)+4n-5#

and we have shown that for #n ge 2#

#a_(n-1)= a_n + 4(n+1)-5#

The final arrangements for the finite induction demonstration are left to the reader.