Solve the following. (10 pts each)
1. Prove P(n) = n2(n + 1)
2. Recurrence relation an = 2n with the initial term a1= 2.
(1) P(n)=n2(n+1)P(n)=n^2(n+1)\\P(n)=n2(n+1)
=n3+n= n^3+n \\=n3+n
Sn=∑P(n)=∑n3+∑nS_n= \sum P(n)=\sum n^3+\sum nSn=∑P(n)=∑n3+∑n
=(n(n+1)2)2+n(n+1)2=n4+n3+2n2+n3+n2+2n4=14n(n+1)(n2+n+2)= (\dfrac{n(n+1)}{2})^2 + \dfrac{n(n+1)}{2}\\ \\ =\dfrac{n^4+n^3+2n^2+n^3+n^2+2n}{4}\\=\dfrac{1}{4}n(n+1)(n^2+n+2)=(2n(n+1))2+2n(n+1)=4n4+n3+2n2+n3+n2+2n=41n(n+1)(n2+n+2)
(2) an=2n a1=2a_n=2n \ \ \ \ \ \ a_1=2an=2n a1=2
This is already a recurrence relation with sum
Sn=∑2n=2n(n+1)2=n(n+1)S_n=\sum 2n=2\dfrac{n(n+1)}{2}=n(n+1)Sn=∑2n=22n(n+1)=n(n+1)
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