R.H.S. = (−1)n−12n(n5+3n4−5n2+3)
=(−1)n−12n(n+1)(n4+2n3−2n2−3n+3).....(1)
by actual division.
P(1) is obiviously true. Assume P(n).
∴ P(n + 1) = P(n) +(−1)n(n+1)6
=(−1)n−12n(n+1)(n4+2n3−2n2−3n+3)+(−1)n(n+1)6
=(−1)n2(n+1)[−(n5+2n4−2n3−3n2+3n)+2(n+1)5]
=(−1)n2(n+1)[2(n5+5n4+10n3+10n2+5n+1)−n5−2n4+2n3+3n2−3n]
=(−1)n2(n+1)[n5+8n4+22n3+23n2+7n+2]
=(−1)n2(n+1)(n+2)
[n4+6n3+10n2+3n+1] ....(2)
Now keeping in view the form (1) the expression within the bracket can be written as
{(n+1)4+2(n+1)3−2(n+1)2−3(n+1)+3}
Hence (2) takes the form of (1) with
n→(n+1)∴P(n)⇒P(n+1)